codeforces-892B Wrath

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 longusing 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;}