Educational Codeforces Round 23

Educational Codeforces Round 23  

A. Treasure Hunt

先判一下是否都是倍数，然后判一下倍数奇偶性

view code

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define pb push_back
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x)  cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

void solve(){
    ____();
    int x1, y1, x2, y2, x, y;
    cin >> x1 >> y1 >> x2 >> y2 >> x >> y;
    int dx = abs(x2 - x1), dy = abs(y2 - y1);
    if(dx%x!=0 or dy%y!=0){
        cout << "NO" << endl;
        return;
    }
    if(((dx/x)&1) == ((dy/y)&1)) cout << "YES" << endl;
    else cout << "NO" << endl;
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

B. Makes And The Product

从小到大排序之后先选前三个，如果后面有和第三个一样的话全部都选进来

然后分情况讨论即可

view code

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define pb push_back
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x)  cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

void solve(){
    int n; sci(n);
    vi A(n);
    for(int &x : A) sci(x);
    sort(all(A));
    vi vec;
    for(int i = 0; i < 3; i++) vec.pb(A[i]);
    for(int i = 3; i < (int) A.size(); i++) if(A[i]==vec.back()) vec.pb(A[i]);
    set<int> S; for(int x : vec) S.insert(x);
    if(S.size()==1){
        cout << 1ll * vec.size() * (vec.size() - 1) * (vec.size() - 2) / 6 << endl;
        return;
    }else if(S.size()==2){
        if(vec[0]==vec[1]) cout << vec.size() - 2 << endl;
        else cout << 1ll * (vec.size() - 1) * (vec.size() - 2) / 2 << endl;
    }else cout << vec.size() - 2 << endl;
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

C. Really Big Numbers

考虑计算差值小于(s)的

显然小于(xle s)的(x)差值必然小于(s)

而且满足条件的(x)不会超过(s)很多，枚举一下(s)附近的(x)即可

view code

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define pb push_back
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x)  cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
int dig(LL x){
    int s = 0;
    while(x) s += x % 10, x /= 10;
    return s;
}
void solve(){
    LL n, s;
    cin >> n >> s;
    LL top = min(s+20000, n);
    LL num = min(s - 1, n);
    for(LL i = s; i <= top; i++) if(i - dig(i) < s) num++;
    cout << n - num << endl;
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

D. Imbalanced Array

考虑每个值对答案的贡献，枚举每个值作为最大值或者最小值在哪些区间出现

我们需要找到对于一个点左边第一个小于等于，右边第一个小于的位置

左边第一个大于等于，右边第一个大于的位置

这个用单调栈维护一下即可

view code

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define pb push_back
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x)  cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e6+7;
int A[MAXN], ls[MAXN], rs[MAXN];

void solve(){
    int n; sci(n);
    for(int i = 1; i <= n; i++) sci(A[i]);
    LL ret = 0;
    stack<int> stk;
    for(int i = 1; i <= n; i++){
        ls[i] = i;
        while(!stk.empty() and A[stk.top()]<=A[i]) ls[i] = ls[stk.top()], stk.pop();
        stk.push(i);
    }
    while(!stk.empty()) stk.pop();
    for(int i = n; i >= 1; i--){
        rs[i] = i;
        while(!stk.empty() and A[stk.top()]<A[i]) rs[i] = rs[stk.top()], stk.pop();
        stk.push(i);
    }
    while(!stk.empty()) stk.pop();
    for(int i = 1; i <= n; i++) ret += 1ll * (rs[i] - i + 1) * (i - ls[i] + 1) * A[i];
    for(int i = 1; i <= n; i++){
        ls[i] = i;
        while(!stk.empty() and A[stk.top()]>=A[i]) ls[i] = ls[stk.top()], stk.pop();
        stk.push(i);
    }
    while(!stk.empty()) stk.pop();
    for(int i = n; i >= 1; i--){
        rs[i] = i;
        while(!stk.empty() and A[stk.top()]>A[i]) rs[i] = rs[stk.top()], stk.pop();
        stk.push(i);
    }
    while(!stk.empty()) stk.pop();
    for(int i = 1; i <= n; i++) ret -= 1ll * (rs[i] - i + 1) * (i - ls[i] + 1) * A[i];
    cout << ret << endl;
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

E. Choosing The Commander

简单的字典树维护

view code

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define pb push_back
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x)  cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e5+7;

int q, tot, ch[MAXN<<4][2], root, sum[MAXN<<4];

void ins(int p, int &rt, int dep){
    if(!rt) rt = ++tot;
    sum[rt] += 1;
    if(dep==-1) return;
    ins(p,ch[rt][p>>dep&1],dep-1);
}

void del(int p, int rt, int dep){
    sum[rt] -= 1;
    if(dep==-1) return;
    del(p,ch[rt][p>>dep&1],dep-1);
}

int query(int p, int l, int rt, int dep){
    if(dep==-1) return 0;
    if(l>>dep&1) return query(p,l,ch[rt][(p>>dep&1)^1],dep-1) + sum[ch[rt][p>>dep&1]];
    else return query(p,l,ch[rt][(p>>dep&1)],dep-1);
}

void solve(){
    sci(q);
    while(q--){
        int tp; sci(tp);
        if(tp==1){
            int p; sci(p);
            ins(p,root,30);
        }else if(tp==2){
            int p; sci(p);
            del(p,root,30);
        }else{
            int p, l;
            sci(p); sci(l);
            printf("%d
",query(p,l,root,30));
        }
    }
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

F. MEX Queries

(l,r)范围很大，离散化之后用线段树维护就好了

主要维护区间置(1)，区间置(0)和区间翻转

view code

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define pb push_back
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define pll pair<LL,LL>
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x)  cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 4e5+7;
struct SgmentTree{
    int l[MAXN<<2], r[MAXN<<2], lm[2][MAXN<<2];
    int tag[MAXN<<2];                                   // 1 reverse, 2 set zero, 3 set one
    #define ls(rt) rt << 1
    #define rs(rt) rt << 1 | 1
    void pushup(int rt){
        lm[0][rt] = min(lm[0][ls(rt)],lm[0][rs(rt)]);
        lm[1][rt] = min(lm[1][ls(rt)],lm[1][rs(rt)]);
    }
    void down(int rt, int tg){
        if(tg==1){
            tag[rt] = 1;
            lm[0][rt] = MAXN;
            lm[1][rt] = l[rt];
        }else if(tg==2){
            tag[rt] = 2;
            lm[0][rt] = l[rt];
            lm[1][rt] = MAXN;
        }else{
            if(!tag[rt]){
                tag[rt] = 3;
                swap(lm[0][rt],lm[1][rt]);
            }else if(tag[rt]==3){
                tag[rt] = 0;
                swap(lm[0][rt],lm[1][rt]);
            }else if(tag[rt]==1){
                tag[rt] = 2;
                lm[0][rt] = l[rt];
                lm[1][rt] = MAXN;
            }else{
                tag[rt] = 1;
                lm[0][rt] = MAXN;
                lm[1][rt] = l[rt];
            }
        }
    }
    void pushdown(int rt){
        if(!tag[rt]) return;
        down(ls(rt),tag[rt]);
        down(rs(rt),tag[rt]);
        tag[rt] = 0;
    }
    void build(int L, int R, int rt = 1){
        l[rt] = L; r[rt] = R;
        lm[0][rt] = l[rt]; lm[1][rt] = MAXN;
        if(l[rt] + 1 == r[rt]) return;
        int mid = (l[rt] + r[rt]) >> 1;
        build(L,mid,ls(rt)); build(mid,R,rs(rt));
        pushup(rt);
    }
    void set(int L, int R, int x, int rt = 1){
        if(L>=r[rt] or l[rt]>=R) return;
        if(L<=l[rt] and r[rt]<=R){
            down(rt,x);
            return;
        }
        pushdown(rt);
        set(L,R,x,ls(rt)); set(L,R,x,rs(rt));
        pushup(rt);
    }
    int query(){ return lm[0][1]; }
}ST;
vector<pair<int,pll> > Q;
int n;
vl vec;
void solve(){
    sci(n);
    Q.resize(n);
    for(auto &p : Q) sci(p.first), scl(p.second.first), scl(p.second.second);
    for(auto p : Q) vec.pb(p.second.first), vec.pb(p.second.second);
    sort(all(vec));
    for(int i = 0, l = vec.size(); i < l; i++){
        vec.pb(vec[i]+1);
        if(vec[i]-1) vec.pb(vec[i]-1);
    }
    vec.pb(1);
    sort(all(vec));
    vec.erase(unique(all(vec)),vec.end());
    ST.build(1,MAXN);
    for(auto &q : Q){
        int l = lower_bound(all(vec),q.second.first) - vec.begin() + 1;
        int r = lower_bound(all(vec),q.second.second) - vec.begin() + 1;
        ST.set(l,r+1,q.first);
        printf("%I64d
",vec[ST.query()-1]);
    }
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}