POJ 3061 subsequence ruler method, a cock dick O (n) algorithm

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9050		Accepted: 3604

Description

A sequence of n positive integers (< n <), each of the them less than or equal 10000, and a positive integer S (S <) 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. The program have to read the numbers N and S, separated by a interval, from the first line. The numbers of the sequence is given in the second line of the "test case", separated by intervals. The input would finish with the end of file.

Output

The the program have to print the result on separate line of the output File.if no answer, print 0.

Sample Input

210 155 1 3 5 10 7 4 9 2 85 111 2 3 4 5

Sample Output

23

#include <cstdio>#include<cmath>#include<cstring>#include<ctime>#include<iostream>#include<algorithm>#include<Set>#include<vector>#include<sstream>#include<queue>#include<typeinfo>#include<fstream>typedefLong Longll;using namespacestd;//freopen ("d.in", "R", stdin);//freopen ("D.out", "w", stdout);#defineSspeed ios_base::sync_with_stdio (0); Cin.tie (0)#defineMAXN 100001Const intinf=0x7fffffff;//infinitely LargeintA[MAXN];intMain () {intT; CIN>>T;  while(t--)    {        intn,s; CIN>>n>>s;  for(intI=0; i<n;i++) {cin>>A[i]; } ll sum=a[0]; intst=0, en=0; intminn=inf;  while(en!=n&&st<=en) {            //cout<<st<< "" <<en<< "" <<sum<<endl;            if(sum>=s) {Minn=min (minn,en-st+1); Sum-=A[st]; St++; }            if(sum<s) {en++; Sum+=A[en]; }        }        if(minn==inf) cout<<"0"<<Endl; Elsecout<<minn<<Endl; }    return 0;}

POJ 3061 subsequence ruler method, a cock dick O (n) algorithm