繼續 count1 read each 長度 long long seq cas sep

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
11
1 2 3 4 5

Sample Output

2
3
題解：求連續子序列和大於等於s的最小長度。可以尺取，若sum<s,左指針右移，若sum>=s右指針左移，當sum<s時左指針繼續右移，滿足條件打擂臺即可。

 1 #include<cstring>
 2
 
 #include<cstdio>
 3 #include<algorithm>
 4 #define Inf 0x3f3f3f3f
 5 #define  ll   long long
 6 using namespace std;
 7 ll str[123400];
 8 int main()
 9 {
10     ll i,j,m,n,t;
11     scanf("%lld",&t);
12     while(t--)
13     {
14         scanf("%lld%lld",&m,&n);
15
 
         for(i=0;i<m;i++)
16         {
17             scanf("%lld",&str[i]);
18         }
19         ll sum=0,cnt=0,cur=0,flag=0,ans=Inf,count1=0;
20         while(cur<m)
21         {
22             while(count1<m&&sum<=n)
23             {
24                 sum+=str[count1++];
25             }
26             if(sum<n)
27                 break;
28                 flag=1;
29             sum-=str[cur];
30             ans=min(ans,count1-cur);
31             cur++;
32         }
33         if(flag)
34         printf("%lld\n",ans);
35         else
36             puts("0");
37     }
38     return 0;
39 
40 }

Sequence（尺取）