【二分】Subsequence

[POJ3061]Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15908		Accepted: 6727

Description

A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the subsequence of consecutive elements of the sequence, the sum of which is greater than or equal to S.

Input

The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The input will finish with the end of file.

Output

For each the case the program has to print the result on separate line of the output file.if no answer, print 0.

Sample Input

2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5

Sample Output

2
3

Source

Southeastern Europe 2006

 

题目大意：问最小区间长度K使得序列中有长度为K的连续子序列相加大于等于S，不够输出0

试题分析：二分序列长度，滑动窗口

 

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
inline int read(){
    int x=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) x=x*10+c-'0';
    return x*f;
}
#define LL long long
const int MAXN=1000001;
const int INF=999999;
const int mod=999993;
int N,S;
int a[1000001];
int tmp;

bool check(int k){
    long long sum=0;
    for(int i=1;i<=k;i++){
        sum+=a[i];
    }
    if(sum>=S) return true;
    for(int i=2;i+k-1<=N;i++){
        sum-=a[i-1];sum+=a[i+k-1];
        if(sum>=S) return true;
    }
    return false;
}

int main(){
    int T=read();
    while(T--){
        N=read(),S=read();long long k=0;
        for(int i=1;i<=N;i++) a[i]=read(),k+=a[i];
        if(k<S) {printf("0
");continue;}
        int l=1,r=N;
        while(l<=r){
            int mid=(l+r)>>1;
            if(check(mid)) r=mid-1;
            else l=mid+1;
        }
        printf("%d
",l);
    }
}