B - Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 128000/64000 KB (Java/Others)
Submit Status
Problem 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
1 #include <stdio.h> 2 #include <queue> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <vector> 7 #include <map> 8 #include <set> 9 #include <ctime> 10 #include <cmath> 11 #include <cctype> 12 using namespace std; 13 typedef long long LL; 14 const int N=1e5+10; 15 const int INF=0x3f3f3f3f; 16 int cas=1,T; 17 int n,a[N],s; 18 int main() 19 { 20 // freopen("1.in","w",stdout); 21 // freopen("1.in","r",stdin); 22 // freopen("1.out","w",stdout); 23 scanf("%d",&T); 24 while(T--) 25 { 26 scanf("%d%d",&n,&s); 27 for(int i=0;i<n;i++) scanf("%d",a+i); 28 int ans=INF; 29 for(int i=0,j=0,sum=0;i<n;i++) 30 { 31 while(sum<s && j<n) 32 { 33 sum+=a[j]; 34 j++; 35 } 36 if(sum>=s) ans=min(ans,j-i); 37 sum-=a[i]; 38 } 39 if(ans==INF) ans=0; 40 printf("%d ",ans); 41 } 42 // printf("time=%.3lf ",(double)clock()/CLOCKS_PER_SEC); 43 return 0; 44 }
原题：POJ3061
题解：
二分或滑动窗口，题解是滑动窗口
二分：枚举每个位置，然后二分找出刚好区间和大于S的另一个位置，记录最小长度
滑动窗口：维护两个下标动态记录长度