Wrath （思维模拟）

Wrath

Hands that shed innocent blood!

There are n guilty people in a line, the i-th of them holds a claw with length L_i. 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 - L_i.

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, ..., L_n (0 ≤ L_i ≤ 10⁹), where L_i 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.

题意

每个人都有一个长度为 li

的武器，相邻的两个人之间距离为 1 ，同一时间所有人使用武器攻击左边的人，问最后存活下来的人数。

 

思路

显然，最右侧的人一定是可以存活下来的。

我们维护一个 cnt

代表右侧延伸到当前位置的武器长度，

• 若 cnt>0

说明当前位置在别人的攻击范围内，否则 ans+1 。更新 cnt 为 max(cnt−1,ai) 看对于 i

• 来说是否可以攻击到更远的位置。

时间复杂度 O(n)

code：

cin会T

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 2e6+10;
int a[maxn],n;
void solve(){
    int cnt = a[n-1],ans = 1;
    for(int i = n-2; i >= 0; i--){
        if(!cnt) ans++;//如果最长延伸长度是0，当前这个人存活
        cnt = max(cnt-1,a[i]);//获得最长延伸长度
    }
    printf("%d\n",ans);
}
int main(){
    scanf("%d",&n);
    for(int i = 0; i < n; i++){
        scanf("%d",&a[i]);
    }
    solve();
    return 0;
}