Codeforces Round #446 (Div. 2) B. Wrath(线段树，RMQ，区间最值)

Description

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

40 1 0 10

output

1

input

20 0

output

2

input

101 1 3 0 0 0 2 1 0 3

output

3

Note

In first sample the last person kills everyone in front of him.

思路

有一群人，给他们分别编号，当钟声响起，他们会用手里的刀杀人，给出了刀的长度，第i个人能杀第j个人的条件是：

j < i 且 j>=i-Li

所以我们利用RMQ或者线段树枚举从1到n之间的i-Li的最小值，然后就可以判断这个人死不死，最后算出来死亡的总人数，再用总数减去就是答案

代码1（RMQ）

#include <cstdio>#include <cstring>#include <cctype>#include <string>#include <set>#include <iostream>#include <stack>#include <cmath>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define ll long longusing namespace std;const int N=1e6+20;int a[N];int minsum[N][20];void RMQ(int num){    for(int j = 1; j < 20; ++j)        for(int i = 1; i <= num; ++i)            if(i + (1 << j) - 1 <= num)            {                minsum[i][j] = min(minsum[i][j - 1], minsum[i + (1 << (j - 1))][j - 1]);            }}int main(){    int n,sum=0;    scanf("%d",&n);    for(int i=1; i<=n; i++)    {        scanf("%d",&a[i]);        minsum[i][0]=i-a[i];    }    RMQ(n);    for(int i=1; i<n; i++)    {        int k=(int)(log(n-(i+1)+1.0)/log(2.0));        int minnum=min(minsum[i+1][k],minsum[n-(1<<k)+1][k]);        if(i>=minnum)sum++;    }    printf("%d\n",n-sum);    return 0;}

代码2（线段树）

#include <cstdio>#include <cstring>#include <cctype>#include <string>#include <set>#include <iostream>#include <stack>#include <cmath>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define ll long longusing namespace std;const int N=1e6+20;int a[N],b[N];#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1int MIN[N<<2];void pushup(int rt){    MIN[rt]=min(MIN[rt<<1],MIN[rt<<1|1]);}void build(int l,int r,int rt){    if(l==r)    {        MIN[rt]=b[l];        return;    }    int m=(l+r)>>1;    build(lson);    build(rson);    pushup(rt);}int querymin(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)        return MIN[rt];    int m=(l+r)>>1;    int ans=inf;    if(L<=m)        ans=min(ans,querymin(L,R,lson));    if(R>m)        ans=min(ans,querymin(L,R,rson));    return ans;}int main(){    int n,sum=0;    scanf("%d",&n);    for(int i=1; i<=n; i++)    {        scanf("%d",&a[i]);        b[i]=i-a[i];    }    build(1,n,1);    for(int i=1; i<n; i++)    {        int minnum=querymin(i+1,n,1,n,1);        if(i>=minnum)sum++;    }    printf("%d\n",n-sum);    return 0;}