POJ 3061 Subsequence

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
 
t题目要求找一个区间和大于S切长度最小的区间 
从头到尾扫一遍 
如果当前和小于s 尾指针加1 
如果当前和大于等于s 删除第一个元素
 

1 #include <cctype>
2 #include <cstdio>
3
4 const int INF=0x3f3f3f3f;
5 const int MAXN=100010;
6
7 int T,n,s,ans;
8
9 int a[MAXN];
10
11 inline void read(int&x) {
12     int f=1;register char c=getchar();
13     for(x=0;!isdigit(c);c=='-'&&(f=-1),c=getchar());
14     for(;isdigit(c);x=x*10+c-48,c=getchar());
15     x=x*f;
16 }
17
18 int hh() {
19     read(T);
20     while(T--) {
21         read(n);read(s);
22         int sum=0;
23         ans=INF;
24         for(int i=1;i<=n;++i) read(a[i]),sum+=a[i];
25         if(s>sum) {printf("0\n");continue;}
26         int head=1,tail=1;
27         sum=a[1];
28         while(head<=tail) {
29             if(sum<s&&tail<n) sum+=a[++tail];
30             else sum-=a[head++];
31             if(sum>=s)
32               if(ans>tail-head+1) ans=tail-head+1;
33         }
34         printf("%d\n",ans);
35     }
36     return 0;
37 }
38
39 int sb=hh();
40 int main(int argc,char**argv) {;}

代码