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.
Example
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：
听说穿女装打acm有buff加成，但是死神达达讨厌女装，所以他要让n名女装大佬自相残杀。 一行中有n名女装大佬，每名女装大佬手中都拿着Li长度的镰刀。当12点钟声响起，每个人都会杀死在他左边的Li个人，该项动作同时进行。死神达达想知道最后存活下来的人数。
题解：
从后往前模拟即可，看代码即可。
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e6+5;
int a[maxn];
int main()
{
int n;
while(cin>>n)
{
for(int i=0;i<n;i++)
cin>>a[i];
int ans=0;
int k=0;
for(int i=n-1;i>=0;i--)
{
if(k==0)
ans++;
k=max(a[i],k-1);
}
cout<<ans<<endl;
}
return 0;
}