Subsequence --二分法

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

23

题意：给出n个数和一个S，求这 n 个数的子序列大于等于 S 的最短序列长度是多少。
题解：求出当 i 为之前面所有数的和。然后从最后一个数开始，找出第一个满足条件的数的位置，求之间的距离更新 ans 值，最后得出最短的 ans 。

#include#includeusing namespace std;int a[100005];int main(){int t;int temp;scanf("%d",&t);while(t--){int n,s;int left,right;int ans=0;scanf("%d%d",&n,&s);a[0]=0;for(int i=1;i<=n;i++){scanf("%d",&a[i]);    a[i]=a[i]+a[i-1];}       if(a[n]=s)       {       flag=1;       break;}}if(flag==1)right=mid;elseleft=mid;}printf("%d\n",right);   }}return 0;}