CF - 797E. Array Queries - dp+有选择地暴力

题目描述：
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.
题意概述：给你一个n个元素的数组; 每个元素都在1..n之间; 然后给你q个询问; 每个询问由p和k构成; 会对p进行 p=p+a[p]+k操作若干次; 你要输出p第一次大于n之后操作了多少次;
解题思路：一种很好的想法是对于小于350的结果先计算出来，直接查询，大于350的结果再逐个dp。还有每次query的结果也可以放入dp内，以免下次继续查询。

AC代码：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 100100
#define lson root << 1
#define rson root << 1 | 1
#define lent (t[root].r - t[root].l + 1)
#define lenl (t[lson].r - t[lson].l + 1)
#define lenr (t[rson].r - t[rson].l + 1)
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
int a[maxn], dp[maxn][350];
int main()
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);long _begin_time = clock();
#endifint n;while (~scanf("%d", &n)){for (int i = 1; i <= n; i++)scanf("%d", &a[i]);for (int i = n; i >= 1; i--)for (int j = 1; j < 350; j++){if (i + a[i] + j > n)dp[i][j] = 1;elsedp[i][j] = dp[i + a[i] + j][j] + 1;}int q;scanf("%d", &q);while (q--){int p, k;scanf("%d%d", &p, &k);if (k < 350)printf("%d\n", dp[p][k]);else{int cnt = 0;while (p <= n){cnt++;p = p + a[p] + k;}printf("%d\n", cnt);}}}
#ifndef ONLINE_JUDGElong _end_time = clock();printf("time = %ld ms.", _end_time - _begin_time);
#endifreturn 0;
}

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