利用尺取法的反向扫描

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<string>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int arr[100010];
int main()
{
	int t, n, m;
	cin >> t;
	for (int i = 0; i < t; i++)
	{
		int sum = 0;
		int zl = 0, zr = 0;//左右下标的状态判断
		cin >> n >> m;
		int l, r;
		l = 0; r = n - 1;//左右下标
		for (int j = 0; j < n; j++)
		{
			cin >> arr[j];
			sum += arr[j];//统计和
		}
		if (sum < m)//特判如果总和小于m，则无解则输出0
		{
			cout << 0 << endl;
		}
		else
		{
			while (1)
			{
				if (zr == 1 && zl == 1)
				{
					break;
				}
				if (sum - arr[l] >= m)//如果减去仍满足，则移动下标
				{
					sum = sum - arr[l];
					l++;
				}
				else
				{
					zl = 1;//左端已不再移动
				}
				if (sum - arr[r] >= m)//同左端
				{
					r--;
					sum = sum - arr[r];
				}
				else
				{
					zr = 1;
				}
			}
			cout << r - l + 1 << endl;//输出长度
		}
	}
}