2018年全国多校算法寒假训练营练习比赛（第五场）

A：逆序数

写过一篇原理，这里不想多说。 传送门

代码：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
const int INF = 0x3f3f3f3f;
const LL mod = 1e9+7;
typedef pair<int,int> pll;
const int N = 100000+10;
int tree[N];
int lowbit(int x)
{
    return x&(-x);
}
void add(int x)
{
    while(x < N)
    {
        tree[x]++;
        x += lowbit(x);
    }
}
int Query(int x)
{
    int ret = 0;
    while(x)
    {
        ret += tree[x];
        x -= lowbit(x);
    }
    return ret;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    LL ans = 0;
    while(n--)
    {
        int tmp;
        cin >> tmp;
        tmp++;
        ans = ans + Query(N-1) - Query(tmp);
        add(tmp);
    }
    cout << ans << endl;
    return 0;
}

B：Big Wate Problem

树状数组裸题

代码：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
const int INF = 0x3f3f3f3f;
const LL mod = 1e9+7;
typedef pair<int,int> pll;
const int N = 100000+10;
LL tree[N];
int n, m;
int lowbit(int x){
    return x&(-x);
}
void add(int x, int c){
    while(x <= n){
        tree[x] += c;
        x += lowbit(x);
    }
}
LL Query(int x){
    LL ret = 0;
    while(x){
        ret += tree[x];
        x -= lowbit(x);
    }
    return ret;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    int t;
    for(int i = 1; i <= n; i++){
        cin >> t;
        add(i,t);
    }
    int x, y;
    while(m--){
        cin >> t >> x >> y;
        if(t == 1){
            add(x,y);
        }
        else {
            cout << Query(y) - Query(x-1) << endl;
        }
    }
    return 0;
}

C：字符串的问题

KMP裸题

注意一点的就是 aaaaaa 输出的是 aaaa， 只要不完全重合位置就好了。

代码：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
const int INF = 0x3f3f3f3f;
const LL mod = 1e9+7;
typedef pair<int,int> pll;
string str;
const int N = 1e6+5;
int nt[N];
int n;
void get_nt(){
    nt[0] = -1;
    int k = -1, j = 0;
    while(j < n){
        if(k == -1 || str[k] == str[j]) nt[++j] = ++k;
        else k = nt[k];
    }
}
bool Kmp(int len){
    for(int i = 1, j = 0; i < n-1; i++){
        while(j > 0 && str[i] != str[j]) j = nt[j];
        if(str[i] == str[j]) j++;
        if(j == len) return true;
    }
    return false;
}
bool pre_suf(int len){
    for(int i = 0; i < len; i++){
        if(str[i] != str[n-len+i]) return false;
    }
    return true;
}
bool check()
{
    for(int i = n-1; i >= 1; i--){
        if(pre_suf(i))
            if(Kmp(i)){
            for(int j = 0; j < i; j++)
                cout << str[j];
            return true;
        }
    }
    return false;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> str;
    n = str.size();
    get_nt();
    if(!check()) cout << "Just a legend
";
    return 0;
}

D：集合问题

DFS搜索， 右转传送门

E：情人节的灯泡

2维树状数组

每次对点(x1,y1)修改的时候 会 影响到区域D,

假设现在查询(x1,y1) ---(x2,y2)。

Query(x2,y2) = A+B+C+D.

Query(x1-1,y1-1) = A

Query(x1-1,y2) = A + C

Query(x2,y1-1) = A + B

然后组合一下就可以得到D的值

代码：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
const int INF = 0x3f3f3f3f;
const LL mod = 1e9+7;
typedef pair<int,int> pll;
const int N = 1e3+10;
int vis[N][N];
int tree[N][N];
int n, m;
int lowbit(int x){
    return x&(-x);
}
void Add(int x, int y, int c){
    for(int i = x; i <= n; i+=lowbit(i)){
        for(int j = y; j <= n; j+=lowbit(j)){
            tree[i][j] += c;
        }
    }
}
int Query(int x, int y){
    int ret = 0;
    for(int i = x; i > 0; i-=lowbit(i)){
        for(int j = y; j > 0; j-=lowbit(j)){
            ret += tree[i][j];
        }
    }
    return ret;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++){
            cin >> vis[i][j];
            if(vis[i][j]) Add(i,j,1);
    }
    int x1, y1, x2, y2, t;
    while(m--){
        cin >> t;
        if(t == 1){
            cin >> x1 >> y1;
            if(vis[x1][y1] == 0) vis[x1][y1] = 1, Add(x1,y1,1);
            else vis[x1][y1] = 0, Add(x1,y1,-1);
        }
        else {
            cin >> x1 >> y1 >> x2 >> y2;
            cout << Query(x2,y2) +  Query(x1-1,y1-1) - Query(x2,y1-1) - Query(x1-1,y2) << endl;
        }
    }
    return 0;
}

F：The Biggest Water Problem

代码：

只要while就好了

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
const int INF = 0x3f3f3f3f;
const LL mod = 1e9+7;
typedef pair<int,int> pll;
int c(int x)
{
    int ret = 0;
    while(x){
        ret += x%10;
        x /= 10;
    }
    return ret;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    while(n >= 10){
        n = c(n);
    }
    cout << n << endl;
    return 0;
}

G：送分啦-QAQ

Fibonacci博弈

代码：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
const int INF = 0x3f3f3f3f;
const LL mod = 1e9+7;
typedef pair<int,int> pll;
set<int> fib;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    fib.insert(3);
    fib.insert(5);
    int a = 3, b = 5, c;
    while(a+b < 1e9+5){
        c = a+b;
        fib.insert(c);
        a = b;
        b = c;
    }
    int n;
    cin >> n;
    if(fib.count(n)){
        cout << "Sha" << endl;
    }
    else cout << "Xian" << endl;
    return 0;
}

H：Tree Recovery

线段树区域修改查询裸题

代码：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
const int INF = 0x3f3f3f3f;
const LL mod = 1e9+7;
typedef pair<int,int> pll;
const int N = 100000+10;
LL tree[N << 2], a[N];
LL lazy[N << 2];
void Update(int rt)
{
    tree[rt] = tree[rt<<1|1] + tree[rt<<1];
}
void PushDown(int rt, int lenl, int lenr)
{
    if(lazy[rt])
    {
        lazy[rt<<1] += lazy[rt];
        lazy[rt<<1|1] += lazy[rt];
        tree[rt<<1] += 1ll*lenl*lazy[rt];
        tree[rt<<1|1] += 1ll*lenr*lazy[rt];
        lazy[rt] = 0;
    }
}
void Add(int L, int R, int C, int l, int r, int rt)
{
    if(L <= l && r <= R)
    {
        lazy[rt] += C;
        tree[rt] += 1ll*(r-l+1)*C;
        return ;
    }
    int m = (l+r)/2;
    PushDown(rt, m-l+1,r-m);
    if(L <= m) Add(L,R,C,lson);
    if(m < R) Add(L,R,C,rson);
    Update(rt);
}
LL Query(int L, int R, int l, int r, int rt)
{
    if(L <= l && r <= R)
        return tree[rt];
    int m = (l+r)/2;
    PushDown(rt,m-l+1,r-m);
    LL ret = 0;
    if(L <= m) ret += Query(L,R,lson);
    if(m < R) ret += Query(L,R,rson);
    return ret;
}
void Build(int l, int r, int rt)
{
    lazy[rt] = 0;
    if(l == r)
    {
        tree[rt] = a[l];
        return ;
    }
    int m = (l+r)/2;
    Build(lson);
    Build(rson);
    Update(rt);
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, m, x, y, z;
    string str;
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
        cin >> a[i];
    Build(1,n,1);
    while(m--){
        cin >> str;
        if(str[0] == 'Q') {
            cin >> x >> y;
            cout << Query(x,y,1,n,1) << endl;
        }
        else {
            cin >> x >> y >> z;
            Add(x,y,z,1,n,1);
        }
    }
    return 0;
}