技術分享 onclick 如果 opened head org .org gre 當前

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題目要求找一個區間和大於S切長度最小的區間 從頭到尾掃一遍 如果當前和小於s 尾指針加1 如果當前和大於等於s 刪除第一個元素

 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) {;}

代碼

POJ 3061 Subsequence