B. Wrath
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

【题意】： n个人排队，第i个人会杀死第j个人当且仅当j < i并且j ≥i-li,问最后有多少个人能幸存下来。

【分析】：从后往前扫，这样就避免了j<i的干扰，记录一下i-Li的最小值，就能判断第j个人是否会被杀死。

//

The i'th person will be alive if min(j - Lj) > i over all j > i.

Consider you know the jth person is alive or not if j > i and you have x = min(j - Lj) over all j > i. If x > i then the ith person will be alive.

And you can update x easily.

【代码】：

#include <bits/stdc++.h>

using namespace std;
typedef  long long ll;
const int inf = 0x7fffffff;
const int N = 1e6+10;
int a[N],b[N],maxn;
int main()
{
    int n,ans;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    
    int maxn = n-a[n];
    ans = 1;
    for(int i=n-1;i>=1;i--)//倒序，求活，即死的反面min(j - Lj) > i over all j > i.
    {
        if(i<maxn)//你的刀还无法触及我
        {
            ans++;//存活√
        }
        maxn = min(maxn,i-a[i]);//想要活？维护一个最小刀长，诶嘿嘿砍不到我~
    }
    printf("%d\n",ans);
    return 0;
}
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