mem inpu ++ numbers reat else number temp using

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22040		Accepted: 9404

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 題意：給定長度為n的數列整數，及整數S。求出總和不小於S的連續子序列的長度的最小值 思路：尺取法 用queue進行維護就可以噠~ acode

#include<iostream>
#include
 
<cstdio>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<string>
#include<vector>
#include<set>
#include<stack>
#include<queue>
#include<map>
#include<cmath>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
#define MAX_N 1000005
#define gcd(a,b) __gcd(a,b)
#define mem(a,x) memset(a,x,sizeof(a))
#define mid(a,b) a+b/2
#define stol(a) atoi(a.c_str())//string to long
int temp[MAX_N];
int main(){
    //std::ios::sync_with_stdio(false);
    //std::cin.tie(0);
//    #ifndef ONLINE_JUDGE
//        freopen("D:\\in.txt","r",stdin);
//        freopen("D:\\out.txt","w",stdout);
//    #else
//    #endif
    int T;
    scanf("%d",&T);
    int N,S;
    while(T--){
        scanf("%d%d",&N,&S);
        for(int i = 0; i < N; i++)
            scanf("%d",&temp[i]);
        int sum = 0;
        queue<int> que;
        int res = inf;
        for(int i = 0; i < N; i++){
            que.push(temp[i]);
            sum += temp[i];
            while(sum >= S){
                res = min(res,(int)que.size());
                sum -= que.front();
                que.pop();
            }
        }
        if(res!=inf)
            printf("%d\n",res);
        else
            printf("0\n");
    }
    return 0;
}

Subsequence POJ - 3061