Naive Operations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 502768/502768 K (Java/Others)
Total Submission(s): 266 Accepted Submission(s): 69

Problem Description
In a galaxy far, far away, there are two integer sequence a and b of length n.
b is a static permutation of 1 to n. Initially a is filled with zeroes.
There are two kind of operations:
1. add l r: add one for al,al+1…ar
2. query l r: query ∑ri=l⌊ai/bi⌋

Input
There are multiple test cases, please read till the end of input file.
For each test case, in the first line, two integers n,q, representing the length of a,b and the number of queries.
In the second line, n integers separated by spaces, representing permutation b.
In the following q lines, each line is either in the form ‘add l r’ or ‘query l r’, representing an operation.
1≤n,q≤100000, 1≤l≤r≤n, there’re no more than 5 test cases.

Output
Output the answer for each ‘query’, each one line.

Sample Input
5 12
1 5 2 4 3
add 1 4
query 1 4
add 2 5
query 2 5
add 3 5
query 1 5
add 2 4
query 1 4
add 2 5
query 2 5
add 2 2
query 1 5

Sample Output
1
1
2
4
4
6

题意 给出序列a，b l-r的权值为 ∑ri=l⌊ai/bi⌋ 给出两种操作。 按要求输出答案。

解题思路：
这是一道简单题。
因为b是1-n的全排列
所以对于每组数据 最多有nlogn次单点更新。
知道这个后，就很好写了。。。。。。。
总复杂度nlognlogn
标程是可持久化分块。。。。 剧毒。

#include <algorithm>
#include <cstdio>
#include <cstring>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
typedef long long LL;
const int MAX=1e5+10;
int b[MAX];
int max1[MAX<<2];
int add1[MAX<<2];
int sum[MAX<<2];
void push_up1(int rt){
    max1[rt]=max(max1[rt<<1],max1[rt<<1|1]);
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void push_dwon(int rt){
    if(add1[rt]){
        add1[rt<<1|1]+=add1[rt];
        add1[rt<<1]+=add1[rt];
        max1[rt<<1]+=add1[rt];
        max1[rt<<1|1]+=add1[rt];
        add1[rt]=0;
    }
    return;
}
void build1(int l,int r,int rt){
    add1[rt]=0;
    max1[rt]=-1e9;
    sum[rt]=0;
    if(l==r){
        add1[rt]=0;
        max1[rt]=-b[l];
        return;
    }
    int m=(l+r)>>1;
    build1(lson);
    build1(rson);
    push_up1(rt);
}
void update1(int L,int R,int l,int r,int rt){
    if(L<=l && r<=R){
        max1[rt]++;
        add1[rt]++;
        return;
    }
    push_dwon(rt);
    int m=(l+r)>>1;
    if(m>=L)
        update1(L,R,lson);
    if(m<R)
        update1(L,R,rson);
    push_up1(rt);
}
void update2(int L,int R,int l,int r,int rt){
    if(l==r){
        while(max1[rt]>=0){
            sum[rt]++;
            max1[rt]-=b[l];
        }
        return;
    }
    push_dwon(rt);
    int m=(l+r)>>1;
    if(m>=L && max1[rt<<1]>=0)
        update2(L,R,lson);
    if(m<R && max1[rt<<1|1]>=0)
        update2(L,R,rson);
    push_up1(rt);
}
int query(int L,int R,int l,int r,int rt){
    if(L<=l && r<=R){
        return sum[rt];
    }
    push_dwon(rt);
    int m=(l+r)>>1;
    int ans=0;
    if(m>=L)
        ans+=query(L,R,lson);
    if(m<R)
        ans+=query(L,R,rson);
    return ans;
}
int main(){
    int n,q;
    while(~scanf("%d %d",&n,&q)){
        for(int i=1;i<=n;i++){
            scanf("%d",&b[i]);
        }
        build1(1,n,1);
        char s[10];
        int l,r;
        while(q--){
            scanf("%s",s);
            if(s[0]=='a'){
                scanf("%d %d",&l,&r);
                update1(l,r,1,n,1);
                update2(l,r,1,n,1);
            }else{
                scanf("%d %d",&l,&r);
                printf("%d\n",query(l,r,1,n,1));
            }
        }
    }
}