Codeforces 892B. Wrath

最新推荐文章于 2019-03-01 20:23:09 发布

shyoldman

最新推荐文章于 2019-03-01 20:23:09 发布

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本文介绍了一种关于多人同时行动的生存者计数问题。在特定的规则下，每个人可以杀死排在其前面且符合条件的人。文章提供了一个算法，用于计算在一轮行动后还活着的人数。

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Hands that shed innocent blood!

There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.

You are given lengths of the claws. You need to find the total number of alive people after the bell rings.

Input

The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.

Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.

Output

Print one integer — the total number of alive people after the bell rings.

Examples

input
4
0 1 0 10

output
1

input
2
0 0

output
2

input
10
1 1 3 0 0 0 2 1 0 3

output
3

Note

In first sample the last person kills everyone in front of him.

题目大意：几个人互相杀人，只要排在他前面而且前面那个人的下标大于等于自己的下标减去L[i]就可以杀，问最后剩多少人

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+7;
int num[maxn];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&num[i]);
	}
	int ans=0;
	int tmp=n+1;
	for(int i=n;i>=1;i--){
		if(i<tmp) ans++;
		if(tmp>=i-num[i]) tmp=i-num[i];
	}
	printf("%d\n",ans);
}