[POJ 3061] Subsequence

最新推荐文章于 2021-08-28 16:54:00 发布

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最新推荐文章于 2021-08-28 16:54:00 发布

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原文链接：http://www.cnblogs.com/TonyNeal/p/poj3061.html

此博客介绍了一个算法挑战，目标是在给定的整数序列中找到长度最小且其和大于等于特定值的连续子序列。通过使用尺取法，实现了一种高效的解决方案。

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Subsequence

TL: 1s, ML: 32Mb

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

Source

Southeastern Europe 2006

【题解】

尺取法应用第二弹，随便搞都能A系列

 1 #include<stdio.h>
 2 using namespace std;
 3 int testc,n,S,a[100010],rr;
 4 int main() {
 5     scanf("%d",&testc);
 6     while(testc--) {
 7         scanf("%d%d",&n,&S);rr=n+1;
 8         for (int i=0;i<n;++i) scanf("%d",&a[i]);
 9         int s=0,t=0,ns=0;
10         for (;;) {
11             while(t<n&&ns<S) ns+=a[t++];
12             if(ns<S) break;
13             if(t-s<rr) rr=t-s;
14             ns-=a[s++];
15         }
16         if(rr>n) rr=0;
17         printf("%d\n",rr);
18     }
19 }