# Codeforces 892 B. Wrath （递推）

## Description

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 ≤ 10^6) — the number of guilty people.

Second line contains n space-separated integers L1, L2, …, Ln (0 ≤ Li ≤ 10^9), 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


## Examples output

1


## 思路

• 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 $\max(cnt-1,a_i)$ 看对于 $i$ 来说是否可以攻击到更远的位置。

## AC 代码

#include<bits/stdc++.h>
#define IO ios::sync_with_stdio(false);\
cin.tie(0);\
cout.tie(0);
using namespace std;
typedef __int64 LL;
const int maxn = 2e6+10;

LL a[maxn],n;

void solve()
{
LL cnt = a[n-1],ans = 1;
for(int i=n-2; i>=0; i--)
{
if(!cnt)ans++;
cnt = max(cnt-1,a[i]);
}
cout<<ans<<endl;
}

int main()
{
IO;
cin>>n;
for(int i=0; i<n; i++)
cin>>a[i];
solve();
return 0;
}