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  • 分块笔记

    1、分块 (区间加法,查询小于x的个数)

    技巧:开vector对区间sort,注意每次暴力更新的区间需要重新把数据放入vector中。

    https://loj.ac/problem/6278

    #include <algorithm>
    #include  <iterator>
    #include  <iostream>
    #include   <cstring>
    #include   <cstdlib>
    #include   <iomanip>
    #include    <bitset>
    #include    <cctype>
    #include    <cstdio>
    #include    <string>
    #include    <vector>
    #include     <stack>
    #include     <cmath>
    #include     <queue>
    #include      <list>
    #include       <map>
    #include       <set>
    #include   <cassert>
    
    using namespace std;
    #define lson (l , mid , rt << 1)
    #define rson (mid + 1 , r , rt << 1 | 1)
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define pb push_back
    #define pq priority_queue
    
    
    
    typedef long long ll;
    typedef unsigned long long ull;
    //typedef __int128 bll;
    typedef pair<ll ,ll > pll;
    typedef pair<int ,int > pii;
    typedef pair<int,pii> p3;
    
    //priority_queue<int> q;//这是一个大根堆q
    //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
    #define fi first
    #define se second
    //#define endl '
    '
    //#define R register
    #define OKC ios::sync_with_stdio(false);cin.tie(0)
    #define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
    #define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
    #define max3(a,b,c) max(max(a,b), c);
    #define min3(a,b,c) min(min(a,b), c);
    //priority_queue<int ,vector<int>, greater<int> >que;
    
    const ll mos = 0x7FFFFFFF;  //2147483647
    const ll nmos = 0x80000000;  //-2147483648
    const int inf = 0x3f3f3f3f;
    const ll inff = 0x3f3f3f3f3f3f3f3f; //18
    const int mod = 1e9+7;
    const double esp = 1e-8;
    const double PI=acos(-1.0);
    const double PHI=0.61803399;    //黄金分割点
    const double tPHI=0.38196601;
    
    
    template<typename T>
    inline T read(T&x){
        x=0;int f=0;char ch=getchar();
        while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
        while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
        return x=f?-x:x;
    }
    
    
    /*-----------------------showtime----------------------*/
    
                const int maxn = 50009;
                ll a[maxn];
                ll sum[maxn];
                vector<ll>s[1009];
                int belong[maxn];
    int main(){
                int n;  scanf("%d", &n);
                for(int i=1; i<=n; i++) scanf("%lld", &a[i]);
                int block = sqrt(n);
                int num = n/block; if(n%block) num++;
                for(int i=1; i<=n; i++)
                {
                    belong[i] = (i-1)/block + 1;
                    s[belong[i]].pb(a[i]);
                }
                for(int i=1; i<=num; i++) sort(s[i].begin(),s[i].end());
                int m = n;
                for(int i=1; i<=m; i++){
                    int op;scanf("%d", &op);
                    if(op == 0){
                        int l,r; ll x;
                        scanf("%d%d%lld", &l, &r, &x);
    
                         int id = belong[l],le = (belong[l]-1)*block+1;
                         if(belong[l] == belong[r]){
                            for(int i=l; i<=r; i++) a[i] += x;
                            s[id].clear();
                            for(int i=le; i<= belong[l]*block; i++) s[id].pb(a[i]);
                            sort(s[id].begin(),s[id].end());
                            continue;
                         }
    
                         for(int i=l; i<=belong[l]*block; i++) a[i] += x;
                         s[id].clear();
                         for(int i=le; i<= belong[l]*block; i++) s[id].pb(a[i]);
                         sort(s[id].begin(),s[id].end());
    
                         for(int i=belong[l]+1; i<=belong[r]-1; i++){
                            sum[i] += x;
                         }
    
                         int rid = belong[r],rle = (rid - 1) * block + 1;
                         for(int i=(rid-1)*block+1; i<=r; i++) a[i]+=x;
                         s[rid].clear();
                         for(int i=rle; i<=min(n, belong[r]*block) ; i++) s[rid].pb(a[i]);
    
                         sort(s[belong[r]].begin(),s[belong[r]].end());
                    }
                    else {
                            int l,r; ll x;
                            scanf("%d%d%lld", &l, &r, &x);
                            int cnt = 0;
                            if(belong[l] == belong[r]){
                                for(int i = l; i<= r; i++){
                                    if(a[i] + sum[belong[i]] < 1ll*x*x) cnt++;
                                }
                            }
                            else {
                                for(int i=l; i<=belong[l]*block; i++)
                                    if(a[i] + sum[belong[i]] < 1ll*x*x) cnt++;
                                for(int i=belong[l]+1; i<=belong[r]-1; i++){
                                    cnt += lower_bound(s[i].begin(),s[i].end(), 1ll*x*x-sum[i]) - s[i].begin();
                                }
                                for(int i=(belong[r]-1)*block+1; i<=r; i++)
                                    if(a[i] + sum[belong[i]] < 1ll*x*x) cnt++;
                            }
                            printf("%d
    ", cnt);
                    }
                }
                return 0;
    }
    View Code

    2、区间加法,区间查询

    开两个标记数组,一个是sum数组,一个是add数组。

    https://loj.ac/problem/6280

    #include <algorithm>
    #include  <iterator>
    #include  <iostream>
    #include   <cstring>
    #include   <cstdlib>
    #include   <iomanip>
    #include    <bitset>
    #include    <cctype>
    #include    <cstdio>
    #include    <string>
    #include    <vector>
    #include     <stack>
    #include     <cmath>
    #include     <queue>
    #include      <list>
    #include       <map>
    #include       <set>
    #include   <cassert>
    
    using namespace std;
    #define lson (l , mid , rt << 1)
    #define rson (mid + 1 , r , rt << 1 | 1)
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define pb push_back
    #define pq priority_queue
    
    
    
    typedef long long ll;
    typedef unsigned long long ull;
    //typedef __int128 bll;
    typedef pair<ll ,ll > pll;
    typedef pair<int ,int > pii;
    typedef pair<int,pii> p3;
    
    //priority_queue<int> q;//这是一个大根堆q
    //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
    #define fi first
    #define se second
    //#define endl '
    '
    //#define R register
    #define OKC ios::sync_with_stdio(false);cin.tie(0)
    #define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
    #define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
    #define max3(a,b,c) max(max(a,b), c);
    #define min3(a,b,c) min(min(a,b), c);
    //priority_queue<int ,vector<int>, greater<int> >que;
    
    const ll mos = 0x7FFFFFFF;  //2147483647
    const ll nmos = 0x80000000;  //-2147483648
    const int inf = 0x3f3f3f3f;
    const ll inff = 0x3f3f3f3f3f3f3f3f; //18
    const int mod = 1e9+7;
    const double esp = 1e-8;
    const double PI=acos(-1.0);
    const double PHI=0.61803399;    //黄金分割点
    const double tPHI=0.38196601;
    
    
    template<typename T>
    inline T read(T&x){
        x=0;int f=0;char ch=getchar();
        while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
        while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
        return x=f?-x:x;
    }
    
    
    /*-----------------------showtime----------------------*/
    
                const int maxn = 50009;
                ll a[maxn];
                ll sum[maxn];
                ll add[maxn];
                int belong[maxn];
    int main(){
                int n;  scanf("%d", &n);
                for(int i=1; i<=n; i++) scanf("%lld", &a[i]);
                int block = sqrt(n);
    
                for(int i=1; i<=n; i++)
                {
                    belong[i] = (i-1)/block + 1;
                    sum[belong[i]] += a[i];
                }
    
                int m = n;
                while(m--){
                    int op; scanf("%d", &op);
                    if(op == 0){
                        int l,r,x;
                        scanf("%d%d%d", &l, &r, &x);
                        if(belong[l] == belong[r]){
                            for(int i=l; i<=r; i++){
                                a[i] += x;
    
                                sum[belong[i]] += x;
                            }
                            continue;
                        }
                        for(int i=l; i <= belong[l]*block; i++){
                            a[i] += x;
                            sum[belong[i]] += x;
                        }
                        for(int i=belong[l]+1; i<=belong[r]-1; i++){
                            sum[i] += block * x;
                            add[i] += x;
                        }
                        for(int i=(belong[r]-1) * block + 1; i<=r; i++){
                            a[i]+=x;
                            sum[belong[i]] += x;
                        }
                    }
                    else {
                        int l,r,x;
                        scanf("%d%d%d", &l, &r, &x);
                        ll ans = 0;
                        if(belong[l] == belong[r]){
                            for(int i=l; i<=r; i++){
                                ans = (ans + a[i] + add[belong[l]])%(x+1);
                            }
                        }
                        else {
                            for(int i=l; i <= belong[l]*block; i++){
                                ans = (ans + a[i] + add[belong[i]])%(x+1);
                            }
                            for(int i=belong[l]+1; i<=belong[r]-1; i++){
                                ans = (ans + sum[i])%(x+1);
                            }
                            for(int i=(belong[r]-1) * block + 1; i<=r; i++){
                                ans = (ans + a[i] + add[belong[i]]) %(x+1);
                            }
                        }
                        printf("%lld
    ", ans);
                    }
    
                }
                return 0;
    }
    View Code

    3、区间开方,区间查询

    https://loj.ac/problem/6281

    由于每个数开方的次数不多,所以暴力去开方就行了,分块记录区间中为0,或1的个数。

    这里 变化sum最好写成

    sum[belong[i]] -= a[i];
    a[i] = sqrt(a[i]);
    sum[belong[i]]+=a[i];

    /*
    int tp = a[j] - sqrt(a[j]);
    a[j] = sqrt(a[j]);
    sum[belong[j]]-=tp;
    */

    上面这个可能有精度问题

    #include <algorithm>
    #include  <iterator>
    #include  <iostream>
    #include   <cstring>
    #include   <cstdlib>
    #include   <iomanip>
    #include    <bitset>
    #include    <cctype>
    #include    <cstdio>
    #include    <string>
    #include    <vector>
    #include     <stack>
    #include     <cmath>
    #include     <queue>
    #include      <list>
    #include       <map>
    #include       <set>
    #include   <cassert>
    
    using namespace std;
    #define lson (l , mid , rt << 1)
    #define rson (mid + 1 , r , rt << 1 | 1)
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define pb push_back
    #define pq priority_queue
    
    
    
    typedef long long ll;
    typedef unsigned long long ull;
    //typedef __int128 bll;
    typedef pair<ll ,ll > pll;
    typedef pair<int ,int > pii;
    typedef pair<int,pii> p3;
    
    //priority_queue<int> q;//这是一个大根堆q
    //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
    #define fi first
    #define se second
    //#define endl '
    '
    //#define R register
    #define OKC ios::sync_with_stdio(false);cin.tie(0)
    #define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
    #define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
    #define max3(a,b,c) max(max(a,b), c);
    #define min3(a,b,c) min(min(a,b), c);
    //priority_queue<int ,vector<int>, greater<int> >que;
    
    const ll mos = 0x7FFFFFFF;  //2147483647
    const ll nmos = 0x80000000;  //-2147483648
    const int inf = 0x3f3f3f3f;
    const ll inff = 0x3f3f3f3f3f3f3f3f; //18
    const int mod = 1e9+7;
    const double esp = 1e-8;
    const double PI=acos(-1.0);
    const double PHI=0.61803399;    //黄金分割点
    const double tPHI=0.38196601;
    
    
    template<typename T>
    inline T read(T&x){
        x=0;int f=0;char ch=getchar();
        while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
        while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
        return x=f?-x:x;
    }
    
    
    /*-----------------------showtime----------------------*/
    
                const int maxn = 50009;
                ll a[maxn];
                ll tot[maxn],sum[maxn];
                int belong[maxn];
    int main(){
                int n;  scanf("%d", &n);
                for(int i=1; i<=n; i++) scanf("%lld", &a[i]);
                int block = sqrt(n);
    
                for(int i=1; i<=n; i++){
                    belong[i] = (i-1)/block + 1;
                    sum[belong[i]] += a[i];
                    if(a[i] == 1 || a[i] == 0) {
                         tot[belong[i]]++;
                    }
                }
                int m = n;
    
                while(m--){
                    int op; scanf("%d", &op);
                    if(op == 0){
                        int l,r,x;
                        scanf("%d%d%d", &l, &r, &x);
                        
                        
                        if(belong[l] == belong[r]){
                            for(int i=l; i<=r; i++){
                                if(a[i] == 1 || a[i] == 0) {
                                        continue;
                                }
                                sum[belong[i]] -= a[i];
                                a[i] = sqrt(a[i]);
                                sum[belong[i]]+=a[i];
                                if(a[i] == 1 || a[i] == 0) {
                                        tot[belong[i]]++;
                                }
                            }
                        }
                        else {
                            for(int i=l; i<=belong[l]*block; i++){
                                if(a[i] <= 1) continue;
                                 sum[belong[i]] -= a[i];
                                a[i] = sqrt(a[i]);
                                sum[belong[i]]+=a[i];
                                if(a[i] == 1 || a[i] == 0) {
                                        tot[belong[i]]++;
                                }
                            }
                            for(int i=belong[l] + 1; i< belong[r]; i++){
                                if(tot[i] == block)continue;
                                for(int j=(i-1)*block+1; j<=i*block; j++){
                                    if(a[j] <= 1) continue;
                                    sum[belong[j]] -= a[j];
                                    a[j] = sqrt(a[j]);
                                    sum[belong[j]]+=a[j];
                                
                                    if(a[j] == 1 || a[j] == 0) {
                                            tot[belong[j]]++;
                                    }
                                }
                            }
                            for(int i=(belong[r]-1)*block+1; i<=r; i++){
                                if(a[i] <= 1) continue;
                                sum[belong[i]] -= a[i];
                                a[i] = sqrt(a[i]);
                                sum[belong[i]]+=a[i];
                                
                                if(a[i] == 1 || a[i] == 0) {
                                        tot[belong[i]]++;
                                }
                            }
                        }
                    }
                    else {
                        int l,r,x;
                        scanf("%d%d%d", &l, &r, &x);
                           r = min(r, n);
                        ll res = 0;
                        if(belong[l] == belong[r]) {
                            for(int i=l; i<=r; i++){
                                res += a[i];
                            }
                        }
                        else {
                            for(int i=l; i<=belong[l]*block; i++){
                                res += a[i];
                            }
                            for(int i=belong[l] + 1; i< belong[r]; i++){
                                    res += sum[i];
                            }
                            for(int i=(belong[r]-1)*block+1; i<=r; i++){
                                res += a[i];
                            }
                        }
                        printf("%lld
    ", res);
                    }
                }
    
                return 0;
    }
    View Code

    4、单点插入,单点查询

    由于是插入操作,所以用动态容器做插入,比如这题用vector,然后就是由于当一个块插入过多时,可以考虑重新分块。

    https://loj.ac/problem/6282

    #include <algorithm>
    #include  <iterator>
    #include  <iostream>
    #include   <cstring>
    #include   <cstdlib>
    #include   <iomanip>
    #include    <bitset>
    #include    <cctype>
    #include    <cstdio>
    #include    <string>
    #include    <vector>
    #include     <stack>
    #include     <cmath>
    #include     <queue>
    #include      <list>
    #include       <map>
    #include       <set>
    #include   <cassert>
    
    using namespace std;
    #define lson (l , mid , rt << 1)
    #define rson (mid + 1 , r , rt << 1 | 1)
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define pb push_back
    #define pq priority_queue
    
    
    
    typedef long long ll;
    typedef unsigned long long ull;
    //typedef __int128 bll;
    typedef pair<ll ,ll > pll;
    typedef pair<int ,int > pii;
    typedef pair<int,pii> p3;
    
    //priority_queue<int> q;//这是一个大根堆q
    //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
    #define fi first
    #define se second
    //#define endl '
    '
    //#define R register
    #define OKC ios::sync_with_stdio(false);cin.tie(0)
    #define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
    #define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
    #define max3(a,b,c) max(max(a,b), c);
    #define min3(a,b,c) min(min(a,b), c);
    //priority_queue<int ,vector<int>, greater<int> >que;
    
    const ll mos = 0x7FFFFFFF;  //2147483647
    const ll nmos = 0x80000000;  //-2147483648
    const int inf = 0x3f3f3f3f;
    const ll inff = 0x3f3f3f3f3f3f3f3f; //18
    const int mod = 1e9+7;
    const double esp = 1e-8;
    const double PI=acos(-1.0);
    const double PHI=0.61803399;    //黄金分割点
    const double tPHI=0.38196601;
    
    
    template<typename T>
    inline T read(T&x){
        x=0;int f=0;char ch=getchar();
        while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
        while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
        return x=f?-x:x;
    }
    
    
    /*-----------------------showtime----------------------*/
    
                const int maxn = 200009;
                int n,block;
                int a[maxn],belong[maxn];
                vector<ll> s[2009];
    
                void build(){
                    int tot = 0;
                    for(int i=1; i<=belong[n]; i++) {
                        for(int j=0; j<s[i].size(); j++){
                            a[++tot] = s[i][j];
                        }
                        s[i].clear();
                    }
                    block = (int)sqrt(tot);
                    
                    for(int i=1; i<=tot; i++){
                        belong[i] = (i-1) / block + 1;
                        s[belong[i]].pb(a[i]);
                    }
                }
    int main(){
                scanf("%d", &n);
                for(int i=1; i<=n; i++)
                    scanf("%d", &a[i]);
                block = (int)sqrt(n);
    
                for(int i=1; i<=n; i++){
                    belong[i] = (i-1) / block + 1;
                    s[belong[i]].pb(a[i]);
                }
                int m = n;
                while(m--){
                    int op; scanf("%d", &op);
                    if(op == 0){
                        ll l,r,x;
                        scanf("%lld%lld%lld", &l, &r, &x);
                        for(int i=1; i<=belong[n]; i++){
                            if(s[i].size() < l) l -= (ll)s[i].size();
                            else {
                                s[i].insert(s[i].begin() + l -1 ,r);
                                if(s[i].size() > 20 * block) build();
                                break;
                            }
                        }
                    }
                    else {
                        ll l,r,x;
                        scanf("%lld%lld%lld", &l, &r, &x);
    
                        for(int i=1; i<=belong[n]; i++){
                            if(s[i].size() < r) r = r - (ll) s[i].size();
                            else {
    
                                printf("%lld
    ", s[i][r-1]);
                                break;
                            }
                        }
                    }
                }
                return 0;
     }
    View Code

    5、询问区间等于c的个数,每次询问把区间每个数改为c

    https://loj.ac/problem/6284

    只要这个块是同一个数,就跳过,否则直接暴力计算并修改。

    开了cg【maxn】表示这个块是否统一,统一为这个c,不是就是-1;

    #include <algorithm>
    #include  <iterator>
    #include  <iostream>
    #include   <cstring>
    #include   <cstdlib>
    #include   <iomanip>
    #include    <bitset>
    #include    <cctype>
    #include    <cstdio>
    #include    <string>
    #include    <vector>
    #include     <stack>
    #include     <cmath>
    #include     <queue>
    #include      <list>
    #include       <map>
    #include       <set>
    #include   <cassert>
    
    using namespace std;
    #define lson (l , mid , rt << 1)
    #define rson (mid + 1 , r , rt << 1 | 1)
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define pb push_back
    #define pq priority_queue
    
    
    
    typedef long long ll;
    typedef unsigned long long ull;
    //typedef __int128 bll;
    typedef pair<ll ,ll > pll;
    typedef pair<int ,int > pii;
    typedef pair<int,pii> p3;
    
    //priority_queue<int> q;//这是一个大根堆q
    //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
    #define fi first
    #define se second
    //#define endl '
    '
    //#define R register
    #define OKC ios::sync_with_stdio(false);cin.tie(0)
    #define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
    #define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
    #define max3(a,b,c) max(max(a,b), c);
    #define min3(a,b,c) min(min(a,b), c);
    //priority_queue<int ,vector<int>, greater<int> >que;
    
    const ll mos = 0x7FFFFFFF;  //2147483647
    const ll nmos = 0x80000000;  //-2147483648
    const int inf = 0x3f3f3f3f;
    const ll inff = 0x3f3f3f3f3f3f3f3f; //18
    const int mod = 1e4+7;
    const double esp = 1e-8;
    const double PI=acos(-1.0);
    const double PHI=0.61803399;    //黄金分割点
    const double tPHI=0.38196601;
    
    
    template<typename T>
    inline T read(T&x){
        x=0;int f=0;char ch=getchar();
        while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
        while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
        return x=f?-x:x;
    }
    
    
    /*-----------------------showtime----------------------*/
    
                const int maxn = 100009;
                int n,block;
                int a[maxn],belong[maxn];
                int cg[maxn];
    
                void push_down(int id){
                    for(int i=(id-1)*block+1; i<=min(n,id*block); i++){
                        a[i] = cg[id];
                    }
                    cg[id] = -1;
               }
    int main(){
                scanf("%d", &n);
                for(int i=1; i<=n; i++)
                    scanf("%d", &a[i]);
                block = (int)sqrt(n);
    
                for(int i=1; i<=n; i++){
                    belong[i] = (i-1) / block + 1;
                    cg[belong[i]] = -1;
                }
                int m = n;
                while(m--){
                    int l,r,c;
                    scanf("%d%d%d", &l, &r, &c);
                    int tot = 0;
                    if(belong [l] == belong [r]){
    
                        int id = belong[l];
                        if(cg[id] >= 0) push_down(id);
                        for(int i=l; i<=r; i++) {
                            if(a[i] == c) tot++;
                            a[i] = c;
                        }
    
                    }
                    else {
                        if(cg[belong[l]]>=0) push_down(belong[l]);
                        for(int i=l; i<=belong[l]*block; i++){
                            if(a[i] == c) tot++;
                            a[i] = c;
                        }
    
                        for(int i=belong[l]+1; i<belong[r]; i++){
                            if(cg[i] >= 0) {
                                if(cg[i] == c) tot += block;
                                cg[i] = c;
                            }
                            else {
                                for(int j = (i-1) * block + 1; j<= i*block; j++){
                                    if(a[j] == c) tot++;
                                    a[j] = c;
                                }
                                cg[i] = c;
                            }
                        }
                        if(cg[belong[r]]>=0)push_down(belong [r]);
                        for(int i=(belong[r]-1)*block + 1; i<=r; i++){
                            if(a[i] == c) tot++;
                            a[i] = c;
                        }
    
                    }
                    printf("%d
    ", tot);
                }
                return 0;
     }
    View Code

    6、不带修改,求区间众数

    https://loj.ac/problem/6285

    块的大小开30很快。

    区间的众数是每个块中的众数中的一个。

    #include <algorithm>
    #include  <iterator>
    #include  <iostream>
    #include   <cstring>
    #include   <cstdlib>
    #include   <iomanip>
    #include    <bitset>
    #include    <cctype>
    #include    <cstdio>
    #include    <string>
    #include    <vector>
    #include     <stack>
    #include     <cmath>
    #include     <queue>
    #include      <list>
    #include       <map>
    #include       <set>
    #include   <cassert>
    
    using namespace std;
    #define lson (l , mid , rt << 1)
    #define rson (mid + 1 , r , rt << 1 | 1)
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define pb push_back
    #define pq priority_queue
    
    
    
    typedef long long ll;
    typedef unsigned long long ull;
    //typedef __int128 bll;
    typedef pair<ll ,ll > pll;
    typedef pair<int ,int > pii;
    typedef pair<int,pii> p3;
    
    //priority_queue<int> q;//这是一个大根堆q
    //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
    #define fi first
    #define se second
    //#define endl '
    '
    //#define R register
    #define OKC ios::sync_with_stdio(false);cin.tie(0)
    #define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
    #define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
    #define max3(a,b,c) max(max(a,b), c);
    #define min3(a,b,c) min(min(a,b), c);
    //priority_queue<int ,vector<int>, greater<int> >que;
    
    const ll mos = 0x7FFFFFFF;  //2147483647
    const ll nmos = 0x80000000;  //-2147483648
    const int inf = 0x3f3f3f3f;
    const ll inff = 0x3f3f3f3f3f3f3f3f; //18
    const int mod = 1e9+7;
    const double esp = 1e-8;
    const double PI=acos(-1.0);
    const double PHI=0.61803399;    //黄金分割点
    const double tPHI=0.38196601;
    
    
    template<typename T>
    inline T read(T&x){
        x=0;int f=0;char ch=getchar();
        while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
        while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
        return x=f?-x:x;
    }
    
    
    /*-----------------------showtime----------------------*/
                const int maxn = 1e5+9;
                int n,blo;
                int a[maxn],val[maxn],be[maxn],cnt[maxn];
                int f[5090][5090];
                vector<int>v[maxn];
                map<int,int>mp;
                void pre(int x){
    
                    memset(cnt, 0, sizeof(cnt));
                    int mx = 0;
                    int res;
                    for(int i=(x-1) * blo + 1; i<=n; i++){
                        int t = be[i];
                        cnt[a[i]]++;
                        if(cnt[a[i]] > mx || (cnt[a[i]] == mx && val[a[i]] < val[res] )){
                            mx = cnt[a[i]];
                            res = a[i];
                        }
                        f[x][t] = res;
                    }
                }
                int query(int x,int l,int r){
                    return upper_bound(v[x].begin(),v[x].end(),r) - lower_bound(v[x].begin(),v[x].end(),l);
                }
    int main(){     
                scanf("%d", &n);
                for(int i=1; i<=n; i++){
                     // scanf("%d", &a[i]);
                     read(a[i]);
                }
    
                // blo = (int)sqrt(n);
                blo = 30;
                int tot = 0;
                for(int i=1; i<=n; i++){
                    be[i] = (i-1)/blo + 1;
                    if(!mp.count(a[i])) mp[a[i]] = ++ tot;
                    int id = mp[a[i]];
                    val[id] = a[i];
                    a[i] = id;
                    v[id].pb(i);
                }   
    
                for(int i=1; i<=be[n]; i++){
                    pre(i);
                }
                int m = n;
                while(m--){
                    int l,r;
                    scanf("%d%d", &l, &r);
                    int mx = 0,ans;
                    for(int i=l; i<=min(r, be[l]*blo); i++){
                        int t = query(a[i], l,r);
                        if(t > mx || (t==mx && val[a[i]] < ans)){
                            ans = val[a[i]];
                            mx = t;
                        }
                    }
    
                    int t = query(f[be[l]+1][be[r]-1],l,r);
                    if(t > mx || (t==mx && val[f[be[l]+1][be[r]-1]] < ans)){
                            ans = val[f[be[l]+1][be[r]-1]];
                            mx = t;
                    }
    
                    if(be[l] < be[r]){
                        for(int i=(be[r]-1)*blo+1; i<=r; i++){
                            int t = query(a[i], l,r);
                            if(t > mx || (t==mx && val[a[i]] < ans)){
                                ans = val[a[i]];
                                mx = t;
                            }
                        }
                    }
                    printf("%d
    ", ans);
                }
                return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/ckxkexing/p/10300122.html
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