zoukankan      html  css  js  c++  java
  • POJ 3061 Subsequence 尺取

    Subsequence
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 14698   Accepted: 6205

    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

     

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<sstream>
    #include<algorithm>
    #include<queue>
    #include<deque>
    #include<iomanip>
    #include<vector>
    #include<cmath>
    #include<map>
    #include<stack>
    #include<set>
    #include<fstream>
    #include<memory>
    #include<list>
    #include<string>
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    #define MAXN 100004
    #define L 31
    #define INF 1000000009
    #define eps 0.00000001
    /*
    尺取
    */
    int a[MAXN], n, s;
    int main()
    {
        int t;
        scanf("%d", &t);
        while (t--)
        {
            scanf("%d%d", &n, &s);
            for (int i = 0; i < n; i++)
                scanf("%d", &a[i]);
            int l = 0,sum = 0,ans = INF;
            for (int r = 0; r < n; r++)
            {
                sum += a[r];
                while (sum >= s)
                {
                    sum -= a[l];
                    ans = min(r - l + 1, ans);
                    l++;
                }
            }
            if (ans != INF)
                printf("%d
    ", ans);
            else
                printf("0
    ");
        }
    }
  • 相关阅读:
    Codeforces Round #197 (Div. 2)
    hdu4499Cannon(搜索)
    poj1054The Troublesome Frog
    hdu4705Y
    hdu1054Strategic Game(树形DP)
    poj2029Get Many Persimmon Trees(最大矩阵和)
    poj3280Cheapest Palindrome(记忆化)
    poj3140Contestants Division
    Spring的AOP机制---- 各类通知总结---- 各类通知总结
    Spring的AOP机制---- AOP最终通知---- AOP最终通知
  • 原文地址:https://www.cnblogs.com/joeylee97/p/6895984.html
Copyright © 2011-2022 走看看