Subsequence_尺取法

最新推荐文章于 2023-12-22 02:23:09 发布

ujn20161222

最新推荐文章于 2023-12-22 02:23:09 发布

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本文链接：https://blog.youkuaiyun.com/ujn20161222/article/details/80157198

本文介绍了一种算法，用于寻找给定整数序列中最短的连续子序列，该子序列的元素之和大于或等于指定的目标值S。文章提供了一个完整的程序实现，包括输入输出示例，帮助读者理解算法的工作原理。

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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

//#include<bits/stdc++.h>
#include<cstdio>
#include<iostream>


using namespace std;
#define ll long long
#define all(x) (x).begin(),x.end()
ll rd(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
#define rep(i,a,n) for(int i=a;i<n;i++)
#define se second
const int inf=0x3f3f3f3f;
const int N=1e5+10;
int a[N];

int main(){
    //ios_base::sync_with_stdio(0);
    //cin.tie(0);cout.tie(0);
    freopen("in.txt","r",stdin);

    int t=rd();
    while(t--){
        int n=rd(),m=rd();
        int tot,l,r;
        tot=r=l=0;
        int ans=inf;
        for(int i=1;i<=n;i++){
            a[i]=rd();
            tot+=a[i],r++;
            while(tot>=m){
                ans=ans>r-l?r-l:ans;
                tot-=a[++l];
            }
        }
        printf("%d\n",ans==inf?0:ans);
    }
}