[codeforces797E]Array Queries

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time limit per test : 2 seconds
memory limit per test : 256 megabytes

$a$ is an array of n positive integers, all of which are not greater than $n$.

You have to process q queries to this array. Each query is represented by two numbers p and k. Several operations are performed in each query; each operation changes $p$ to $p + a_p + k$. There operations are applied until $p$ becomes greater than $n$. The answer to the query is the number of performed operations.

Input

The first line contains one integer $n (1 ≤ n ≤ 100000)$.
The second line contains $n$ integers — elements of $a (1 ≤ ai ≤ n$ for each $i$ from $1$ to $n)$.
The third line containts one integer $q (1 ≤ q ≤ 100000)$.
Then $q$ lines follow. Each line contains the values of $p$ and $k$ for corresponding query $(1 ≤ p, k ≤ n)$.
Output

Print $q$ integers, ith integer must be equal to the answer to ith query.

Example
Input

3
1 1 1
3
1 1
2 1
3 1

Output

2
1
1

Note

Consider first example:
In first query after first operation p = 3, after second operation p = 5.
In next two queries p is greater than n after the first operation.

题意：
给一个数组a(a[i]<=n)，长度为n(<=10w)，q(<=10w)个询问，每次询问两个数字p,k，问最多能做几次操作
操作：p=p+a[p]+k（当p>n之后就不能做了）

题解：
对于 $k&gt;\sqrt{n}$的情况，我们发现操作个数不会超过 $\sqrt{n}$，那就暴力即可
对于 $k&lt;\sqrt{n}$的情况，我们用 $n\sqrt{n}$的时间预处理出所有结果即可

#include<bits/stdc++.h>
#define LiangJiaJun main
using namespace std;
int f[404][100004];
int n,q,a[100004];
void ex(int k,int p){
    if(p>n)return ;
    if(f[k][p])return ;
    ex(k,p+k+a[p]);
    if(p+k+a[p]<=n)f[k][p]=f[k][p+k+a[p]]+1;
    else f[k][p]=1;
}
int bruteforce(int k,int p){
    int ans=0;
    while(p<=n){
        ++ans;
        p=p+k+a[p];
    }
    return ans;
}
int LiangJiaJun(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    memset(f,0,sizeof(f));
    for(int i=1;i<=sqrt(n);i++){
        for(int j=1;j<=n;j++){
            if(!f[i][j])ex(i,j);
        }
    }
    scanf("%d",&q);
    while(q--){
        int p,k;
        scanf("%d%d",&p,&k);
        if(k<=sqrt(n)){
            printf("%d\n",f[k][p]);
        }
        else{
            printf("%d\n",bruteforce(k,p));
        }
    }
    return 0;
}