Wrath
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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：第i个人可以开枪杀死它之前的a[i]个人。最后求出活下来的人的个数。
思路：模拟。。。（错了一万次。。。）
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<map>
#define ll long long
using namespace std;
ll a[1000010];
int main()
{
ll n;
scanf("%lld",&n);
for(ll i=1; i<=n; i++)
scanf("%lld",&a[i]);
if(a[n]>=n-1)
{
cout<<1<<endl;
return 0;
}
ll l=n,r=n,ans=0;
for(ll j=n; j>=1; j--)
{
if(a[j]>0)
{
ll x=min(a[j],j);
l=min(l,j-x);
if(l<=0)
l=1;
if(l<r)
ans+=(min(j,r)-l);
r=l;//cout<<l<<" "<<j<<" "<<ans<<endl;
}
}
cout<<n-ans<<endl;
}