POJ3061 Subsequence 尺取法_acm a sequence of n positive integers (10 < n

本文链接：https://blog.youkuaiyun.com/qq_31036127/article/details/106147195

本文探讨了在给定正整数序列与目标值的情况下，寻找最短连续子序列使其和大于等于目标值的问题。通过双指针技巧，即尺取法，实现高效算法解决此类问题。输入包括测试案例数量、序列长度、目标值及序列元素，输出为符合条件的最短子序列长度。

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链接：https://ac.nowcoder.com/acm/problem/107658
来源：牛客网

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.
输入描述:
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.
输出描述:
For each the case the program has to print the result on separate line of the output file.if no answer, print 0.
示例1
输入
复制

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

输出
复制

2
3

尺取法的板子题
用两个指针(lef, rig)维护一段连续区间
sum维护当前区间的和,当sum<m时,我们就应该向右扩展区间
相反,当sum>=m时,我们就应该收缩左区间

#define debug
#ifdef debug
#include <time.h>
#include "/home/majiao/mb.h"
#endif


#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <stdio.h>
#include <math.h>
#define MAXN ((int)1e5+7)
#define ll long long int
#define INF (0x7f7f7f7f)
#define QAQ (0)

using namespace std;

#ifdef debug
#define show(x...) \
do { \
	cout << "\033[31;1m " << #x << " -> "; \
	err(x); \
} while (0)
void err() { cout << "\033[39;0m" << endl; }
#endif

template<typename T, typename... A>
void err(T a, A... x) { cout << a << ' '; err(x...); }

int n, m, Q, K, a[MAXN];

int main() {
#ifdef debug
	freopen("test", "r", stdin);
	clock_t stime = clock();
#endif
	scanf("%d ", &Q);
	while(Q--) {
		scanf("%d %d ", &n, &m);
		for(int i=1; i<=n; i++) scanf("%d ", a+i);
		int lef = 1, rig = 0, ans = INF, sum = 0;
		while(lef <= n) {
			//不断扩展右端点
			while(rig+1<=n && sum<m) sum += a[++rig];
			if(sum >= m) 
				ans = min(ans, rig-lef+1);
			sum -= a[lef]; //收缩左端点
			lef ++;
		}
		//注意没有答案的时候输出0即可
		printf("%d\n", ans==INF ? 0 : ans);
	}

#ifdef debug
	clock_t etime = clock();
	printf("rum time: %lf 秒\n",(double) (etime-stime)/CLOCKS_PER_SEC);
#endif 
	return 0;
}