Codeforces Round #449 (Div. 1) E. Welcome home, Chtholly

E. Welcome home, Chtholly
time limit per test3 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
— I… I survived.

— Welcome home, Chtholly.

— I kept my promise…

— I made it… I really made it!

After several days of fighting, Chtholly Nota Seniorious miraculously returned from the fierce battle.

As promised, Willem is now baking butter cake for her.

However, although Willem is skilled in making dessert, he rarely bakes butter cake.

This time, Willem made a big mistake — he accidentally broke the oven!

Fortunately, Chtholly decided to help him.

Willem puts n cakes on a roll, cakes are numbered from 1 to n, the i-th cake needs ai seconds of baking.

Willem needs Chtholly to do m operations to bake the cakes.

Operation 1: 1 l r x

Willem asks Chtholly to check each cake in the range [l, r], if the cake needs to be baked for more than x seconds, he would bake it for x seconds and put it back in its place. More precisely, for every i in range [l, r], if ai is strictly more than x, ai becomes equal ai - x.

Operation 2: 2 l r x

Willem asks Chtholly to count the number of cakes in the range [l, r] that needs to be cooked for exactly x seconds. More formally you should find number of such i in range [l, r], that ai = x.

Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105).

The second line contains n integers, i-th of them is ai (1 ≤ ai ≤ 105).

The next m lines are the m operations described above. It is guaranteed that 1 ≤ l ≤ r ≤ n and 1 ≤ x ≤ 105.

Output
For each operation of the second type, print the answer.

Examples
input
5 6
1 5 5 5 8
2 2 5 5
1 2 4 3
2 2 5 2
2 2 5 5
1 3 5 1
2 1 5 1
output
3
3
0
3
input
7 7
1 9 2 6 8 1 7
2 1 7 1
2 2 5 2
1 4 7 7
2 2 4 2
1 3 4 5
2 3 3 3
2 3 7 2
output
2
1
1
0
1
input
8 13
75 85 88 100 105 120 122 128
1 1 8 70
2 3 8 30
1 3 8 3
2 2 5 15
1 2 4 10
2 1 5 5
1 2 7 27
2 1 5 5
1 3 7 12
1 1 7 4
2 1 8 1
1 4 8 5
2 1 8 1
output
1
2
3
4
5
6
题意：给你一个数组，有两种操作，1：把l，r之间大于x的数字减去x，2：询问l到r之间x的数量。
做法：对于第一种操作可以把它分成两类1：全部的数减去x，然后小于1的加上x，2：大于x的数减去x；对于一个集合来说，用L来维护这个区间的要减去的数，R维护这个区间最大的数，
并且用并查集维护相等的点，然后如果2*x>R-L,那么久执行第1类操作，把小于L+x的数加上x，然后L加上x，如果x>R-L,那执行第二类操作，把大于x的数减去x。这样维护的话最多进过R-L次操作，所有的数就变成相同的数了。
具体做法：分块，每块用并查集维护，每次把相等的数连起来，用a[x][i]维护第x块里面值为i的数有多少，然后查询的时候第i块值为y的个数就是a[x][L[x]+y]，可以用head[x][i],来维护第x块值为i的并查集的根，每次展开空间的时候只需要把a[x][num[i]]和head[x][num[i]]重置就可以保证L[x]到R[x]区间内的a，head都是重置的。

#include<bits/stdc++.h>using namespace std;const int N = 1e5+1000;const int B = 333;int fa[N],a[B][N],head[B][N],L[N],R[N],pos[N],num[N];int Find(int x){return x == fa[x]?x:fa[x]=Find(fa[x]);}void pushdown(int x){    for(int i = x*B;i < x*B+B;i ++){        num[i] = pos[Find(i)];        a[x][num[i]] = 0;        head[x][num[i]] = -1;    }}void update(int x){    R[x] = 0;    for(int i = x*B;i < x*B+B;i ++){        R[x] = max(R[x],num[i]);        if(head[x][num[i]] == -1){            head[x][num[i]] = i;            pos[i] = num[i];            fa[i] = i;        }        else fa[i] = head[x][num[i]];        a[x][num[i]]++;    }}void mergee(int x,int p,int q){    if(head[x][p] != -1){        if(head[x][q] == -1){            head[x][q] = head[x][p];            a[x][q] = a[x][p];            pos[head[x][p]] = q;        }        else{            fa[head[x][p]] = head[x][q];            a[x][q] += a[x][p];        }    }}void solve(int l,int r,int x){    int ll = l/B,rr = r/B;    if(ll == rr){        pushdown(ll);        for(int i = l;i <= r;i ++){            if(num[i]>L[ll]+x) num[i] -= x;        }        update(ll);    }    else{        pushdown(ll),pushdown(rr);        for(int i = l;i < ll*B+B;i ++){            if(num[i] > L[ll]+x) num[i] -= x;        }        for(int i = rr*B;i <= r;i ++){            if(num[i] > L[rr]+x) num[i] -= x;        }        update(ll),update(rr);        for(int i = ll+1;i < rr;i ++){            if(R[i] > 2*x+L[i]){                for(int j = L[i]+1;j <= L[i]+x;j ++) mergee(i,j,j+x);                L[i]+=x;            }            else if(R[i] > x+L[i]){                for(int j = R[i];j > L[i]+x;j --){                    mergee(i,j,j-x);                }                R[i] = x+L[i];            }        }    }}int query(int l,int r,int x){    int ll = l/B,rr = r/B;    if(ll == rr){        pushdown(ll);        int ans = 0;        for(int i = l;i <= r;i ++)            if(L[ll]+x == num[i]) ans ++;        update(ll);        return ans;    }    else{        int ans = 0;        for(int i = l;i < ll*B+B;i ++){            if(L[ll]+x == pos[Find(i)]) ans ++;        }        for(int i = rr*B;i <= r;i ++){            if(L[rr]+x == pos[Find(i)]) ans ++;        }        for(int i = ll+1;i < rr;i ++)            if(L[i]+x <= R[i]) ans += a[i][L[i]+x];        return ans;    }}int main(){    int n,m;    memset(head,-1,sizeof(head));    scanf("%d %d",&n,&m);    for(int i = 1;i <= n;i ++) scanf("%d",&num[i]);    for(int i = 0;i <= n/B;i ++)        update(i);    for(int i = 1;i <= m;i ++){        int op,l,r,x;        scanf("%d %d %d %d",&op,&l,&r,&x);        if(op == 1) solve(l,r,x);        else printf("%d\n",query(l,r,x));    }    return 0;}