2019 GDUT 新生专题Ⅰ A题_思路 of n positive integers (10 < n

本文链接：https://blog.youkuaiyun.com/weixin_45794203/article/details/103952800

本文介绍了一个尺取法的应用案例，旨在从给定的整数序列中找到连续子序列的最小长度，使得其元素之和大于或等于特定阈值。通过使用尺取法，我们能够有效地解决这一问题，代码示例展示了如何实现这一算法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A 尺取/POJ3061

题目
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

题目大意是在所给的N个元素中，找出连续元素的子序列的最小长度并使他们的和大于等于S。

思路
简单的尺取模板即可。

代码

#include <cstdio>
#include <cstring>
const int inf=100005;
int a[100005];
int main(){
	int n,s,t;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&s);
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);
		}
		int p1=0,p2=0;
		int sum=a[0],min=inf;
		while(p2<n){
			if(sum>=s){
				if(min>p2-p1+1) min=p2-p1+1;
				sum-=a[p1++];
			}
			if(sum<s){
				p2++;
				sum+=a[p2];
			}
		}
		if(min==inf) printf("0\n");
		else
		printf("%d\n",min);
	}
	return 0;
}

题目链接