Codeforces-446(Div.2)-B-Wrath--(线段树区间更新）

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 L_i. 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 - L_i.

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, ..., L_n (0 ≤ L_i ≤ 10⁹), where L_i 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：每个人有长度为ai的武器，可以杀死他左边在攻击范围之内的人

思路：没有想到简单一点的办法，用线段树区间更新了一下，最后查询活着获得人数

AC代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define maxn 1000100
using namespace std;
int n;
struct node
{
    int l,r;
    int tag;//为1代表活着，为0代表被杀死了
}tree[maxn*4];
void build(int root,int l,int r)
{
    tree[root].l=l;
    tree[root].r=r;
    tree[root].tag=1;
    if(l==r)
        return;
    int mid=(l+r)>>1;
    build(root<<1,l,mid);
    build(root<<1|1,mid+1,r);
}
void pushdown(int root)
{
    if(tree[root].tag==0)
    {
        tree[root<<1].tag=0;
        tree[root<<1|1].tag=0;
        tree[root].tag=1;
    }
}
void update(int root,int l,int r)
{
    int ll=tree[root].l;
    int rr=tree[root].r;
    if(l<=ll&&rr<=r){
        tree[root].tag=0;
        return;
    }
    pushdown(root);
    int mid=(ll+rr)>>1;
    if(l<=mid) update(root<<1,l,r);
    if(r>mid) update(root<<1|1,l,r);
}
int query(int root,int p)
{
    int ll=tree[root].l;
    int rr=tree[root].r;
    if(ll==rr)
        return tree[root].tag;
    pushdown(root);
    int mid=(ll+rr)>>1;
    if(p<=mid) return query(root<<1,p);
    else return query(root<<1|1,p);
}
int main()
{
    while(~scanf("%d",&n))
    {
        int x;
        build(1,1,n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(x==0||i==1) continue;
            int mi=max(1,i-x);
            update(1,mi,i-1);
        }
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            if(query(1,i))
                cnt++;
        }
        printf("%d\n",cnt);
    }
    return 0;
}