POJ----3061

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

210 155 1 3 5 10 7 4 9 2 85 111 2 3 4 5

Sample Output

2
3
题目的意思就是：输入n，m，然后给你n个数，这n个数中，连续几个数加起来>m 并且找出最小的个数。
题解：这一道题就是用for循环遍历，然后判断是否>m 并且记录个数，并且与Min比较找到最小的值，然后减去一个数，继续判断
#include <stdio.h>#include <string.h>#include <iostream>#include <math.h>#include<stack>#include<vector>#include<queue>#include<algorithm>using namespace std;int a[110000];int main(){    int n, m, i,  T, sum, l, r;    scanf("%d", &T);    while(T--)    {        scanf("%d%d", &n,&m);        for(i=0;i<n;i++)            scanf("%d", &a[i]);        sum=r=l=0;        int num=m+1;        for(i=0;i<n;i++)        {            while(sum<m && r<n)            {                sum+=a[r];                r++;            }            if(sum<m)                break;            num=min(num, r-l);            sum-=a[l];            l++;        }        if(num>m)            printf("0\n");        else            printf("%d\n", num);    }    return 0;}