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
经典的尺取法，用来判断一段连续的序列的加和，自然而然的想到尺取法。
以前对尺取法理解的不够到位，没有明白尺取法的精妙所在，如今知道了，不得不赞佩其什么的复杂度。
当第一次发现成立的连续段时，不要退出，继续推进，可以得到所有的满足条件的连续段，进而得到最短的那个段。
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 #include<math.h> 5 #include<algorithm> 6 #include<queue> 7 #include<stack> 8 #include<deque> 9 #include<iostream> 10 using namespace std; 11 int con[100009]; 12 int main() 13 { 14 int i,p,j; 15 int n,s,t,head,tail,flag,mid; 16 int big; 17 long long sum=0; 18 while(scanf("%d",&t)!=EOF) 19 { 20 for(i=1; i<=t; i++) 21 { 22 scanf("%d%d",&n,&s); 23 for(j=0; j<n; j++) 24 scanf("%d",&con[j]); 25 big=1000000000; 26 27 head=tail=0; 28 flag=0; 29 sum=con[0]; 30 while(1) 31 { 32 if(head<tail||head>=n||tail>=n||head<0||tail<0) 33 break; 34 if(sum<s) 35 { 36 head++; 37 sum+=con[head]; 38 } 39 else if(sum>=s) 40 { 41 mid=head-tail+1; 42 if(mid<big) 43 big=mid; 44 sum-=con[tail]; 45 tail++; 46 47 } 48 } 49 if(big==1000000000) 50 printf("0 "); 51 else 52 printf("%d ",big); 53 } 54 } 55 56 return 0; 57 }