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POJ - 3061 Subsequence(最短子序列和大于等于某值的序列长度)

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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

题目链接:http://poj.org/problem?id=3061

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题意:

分析:

AC代码1:

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<math.h>
 4 #include<queue>
 5 #include<algorithm>
 6 #include<time.h>
 7 #include<stack>
 8 using namespace std;
 9 #define N 1200000
10 #define INF 0x3f3f3f3f
11 
12 int dp[N];
13 int a[N];
14 
15 int main()
16 {
17     int n,m,T,i;
18 
19     scanf("%d", &T);
20 
21     while(T--)
22     {
23         scanf("%d %d", &n, &m);
24         memset(dp,0,sizeof(dp));
25 
26         for(i=1;i<=n;i++)
27         {
28             scanf("%d", &a[i]);
29             dp[i]=dp[i-1]+a[i];
30         }
31 
32         int minn=INF;
33         for(i=1;i<=n;i++)
34             if(dp[n]-dp[i-1]>=m)
35             {
36                 int ans=lower_bound(dp,dp+n,dp[i-1]+m)-dp;///二分函数
37                 minn=min(ans-i+1,minn);
38             }
39 
40       if(minn==INF)
41         printf("0\n");
42       else
43         printf("%d\n", minn);
44     }
45     return 0;
46 }

 

AC代码2:

 

POJ - 3061 Subsequence(最短子序列和大于等于某值的序列长度)

原文:http://www.cnblogs.com/weiyuan/p/5788033.html

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