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
Copy
4 0 1 0 10
Output
Copy
1
Input
Copy
2 0 0
Output
Copy
2
Input
Copy
10 1 1 3 0 0 0 2 1 0 3
Output
Copy
3
Note
In first sample the last person kills everyone in front of him.
题目链接：http://codeforces.com/contest/892/problem/B
题意：有n个人，每个人都有一个伤害范围是ai的，只能是后面一个杀前面一个而且前面的人在后面的人的伤害范围内。问最后还剩多少人。
从最后一个人（肯定存活）从前开始枚举，每次取伤害范围最大的。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
int a[maxn];
int main()
{
int n;
while(~scanf("%d", &n))
{
for(int i = 1; i <= n; i++)
scanf("%d", a+i);
int ans = a[n];
int sum = 1;
for(int i = n-1; i > 0; i--)
{
if(ans == 0)
sum ++;
ans = max(ans-1, a[i]); // -1是跟前面的一个人比较伤害范围
}
printf("%d\n", sum);
}
return 0;
}