codeforces 797E Array Queries

E. Array Queries

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 + ap + 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.

题目分析:

   直接按照题目描述去做肯定会超时，需要用到dp来优化速度;  

  d[p][k] 表示 p, k需要的次数 状态转移如下：

                              d[p][k] = d[p+a[p]+k][k] + 1; 

  所有的k值保存不下， 这里我保存k值小于200的dp值, d[p][k]  （k < 200）, 当k大于200时直接模拟;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = (int)1e5+5;
int n, q, k, p;
int d[maxn][205], a[maxn];

int solve(int p, int k){
    if(p > n) return 0;
    else if(k < 200){
        if(d[p][k] != -1) return d[p][k];
        else {
            d[p][k] = solve(p+a[p]+k, k)+1;
            return d[p][k];
        }
    } else {
        return solve(p+a[p]+k, k)+1;
    }
}

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    scanf("%d", &q);
    int ans = 0;
    memset(d, -1, sizeof(d));
    for(int i = 0; i < q; ++i){
        scanf("%d%d", &p, &k);
        ans = solve(p, k);
        printf("%d\n", ans);
    }
}