poj3061 Subsequence

//O(nlogn)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX_N = 1e5;
int n, S;
int a[MAX_N], sum[MAX_N + 1];

void solve()
{
	memset(sum, 0, sizeof(sum));
	// 计算 sum
	for (int i = 0; i < n; i++)
	sum[i + 1] = sum[i] + a[i];
	
	if (sum[n] < S)
	{
		cout << 0 << endl;
		return;
	} 
	
	int res = n;
	for (int s = 0; sum[s] + S <= sum[n]; s++)
	{
		// 利用二分搜索法求出t
		int t = lower_bound(sum + s, sum + n, sum[s] + S) - sum - s; 
		res = min(res, t);
	}
	cout << res << endl;
}

int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		cin >> n >> S;
		for (int i = 0; i < n; i++) cin >> a[i];
		solve(); 
	}
	return 0;
}

//优化：O(n)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX_N = 1e5;
int n, S;
int a[MAX_N], sum[MAX_N + 1];
void solve()
{
	int res = n + 1;
	int s = 0, t = 0, sum = 0;  // 初始化
	for (; ; )
	{
		while (t < n && sum < S) // 只要 t没有超出范围， 且sum没达到S，就不断将sum增加a[t]，并且将t增加1 
		{
			sum += a[t++];
		}
		if (sum < S) break; // 若所有数加起来，仍小于S，则此后不可能再有res的更新，可以退出循环，否则更新res 
		
		res = min(res, t - s);
		sum -= a[s++]; //  将sum减去as，s增加1后，回到s不断增加at，t++的循环 
	}
	if (res > n) res = 0;
	cout << res << endl;
}
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		cin >> n >> S;
		for (int i = 0; i < n; i++) cin >> a[i];
		solve(); 
	}
	return 0;
}