POJ---3061 Subsequence[대기열 구문 구간 및]

Subsequence
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 6746
 
Accepted: 2465
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
 
 
 
 
제목: 순서에 따라 그룹의 수를 정하고 이 그룹의 트리와 m보다 큰 최단 서열을 구한다.
 
 
문제의 실질은 하나의 대기열을 유지하고 방향이 왼쪽에서 오른쪽으로 향하며 대기열의 모든 수치와 m보다 크고 가장 짧은 길이를 확보하는 것이다.
 
 
code:

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<algorithm>

 4 using namespace std;

 5 

 6 int data[100100];

 7 int n,m;

 8 

 9 int fun()

10 {

11     int i=0,j;

12     int minx=0x7fffffff;

13     int sum=0;

14     for(j=0;j<n;)

15     {

16         while(sum<m&&j<n)

17             sum+=data[j++];

18         while(sum>=m)

19         {

20             minx=min(minx,j-i);

21             sum-=data[i++];            

22         }

23     }

24     if(minx==0x7fffffff)

25         return 0;

26     return minx;

27 }

28 

29 int main()

30 {

31     int t;

32     int i;

33     scanf("%d",&t);

34     while(t--)

35     {

36         scanf("%d%d",&n,&m);

37         for(i=0;i<n;i++)

38             scanf("%d",&data[i]);

39         printf("%d
",fun());

40     }

41     return 0;

42 }