Subsequence

Time Limit: 1000MS Memory Limit: 65536K

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

2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5

Sample Output

2
3

思路：

就是先用sum存到这个点的时候的总和值，然后因为数值要大于题目所给的m值，所以sum[i] + m如果有能够大于这个的第一个值，那么肯定就是i时最小的长度了，这个时候将两个位置相减就行了。

第一种做法：

#include <iostream>
#include <algorithm>
using namespace std;
int main() {
	int k, n, m;
	scanf("%d", &k);
	while (k--) {
		scanf("%d %d", &n, &m);
		int a[100010] = {0}, sum[100010] = {0};
		for (int i = 0; i < n; i++) {
			scanf("%d", &a[i]);
			sum[i + 1] = sum[i] + a[i];
		}
		if (sum[n] < m) {
			cout << "0\n";
			continue;
		} 
		int ans = n;
		for (int i = 0; sum[i] + m <= sum[n]; i++) {
			int t = lower_bound(sum + i, sum + n, sum[i] + m) - sum;
			ans = min(ans, t - i);
		}
		cout << ans << endl;
	}
	return 0;
}

第二种做法：

和第一种思路没啥区别，就是省了一个数组。

#include <iostream>
#include <algorithm>
using namespace std;
int main() {
	int n, m, k;
	scanf("%d", &k);
	while (k--) {
		int a[100010];
		scanf("%d %d", &n, &m);
		for (int i = 0; i < n; i++) scanf("%d", &a[i]);
		int s = 0, t = 0, sum = 0, ans = n + 1;
		while (1) {
			while (t < n && sum < m) sum += a[t++];
			if (sum < m) break;
			ans = min(ans, t - s);
			sum -= a[s++];
		}
		if (ans > n) cout << "0\n";
		else cout << ans << endl;
	}
	return 0;
}