zoukankan      html  css  js  c++  java
  • POJ 3663 Costume Party (快速排序)

    Costume Party

    Time Limit: 1000MS

     

    Memory Limit: 65536K

    Total Submissions: 9132

     

    Accepted: 3552

    Description

    It's Halloween! Farmer John is taking the cows to a costume party, but unfortunately he only has one costume. The costume fits precisely two cows with a length of (1 ≤ S ≤ 1,000,000). FJ has N cows (2 ≤ N ≤ 20,000) conveniently numbered 1..N; cow i has length Li (1 ≤ Li ≤ 1,000,000). Two cows can fit into the costume if the sum of their lengths is no greater than the length of the costume. FJ wants to know how many pairs of two distinct cows will fit into the costume.

    Input

    * Line 1: Two space-separated integers: N and S
    * Lines 2..N+1: Line i+1 contains a single integer: Li

    Output

    * Line 1: A single integer representing the number of pairs of cows FJ can choose. Note that the order of the two cows does not matter.

    Sample Input

    4 6

    3

    5

    2

    1

    Sample Output

    4

    Source

    USACO 2008 January Bronze

     解题报告:这道题就是让求两牛之间的距离小于等于s的个数,直接遍历找的话TEL,先利用快排从小到大排序, 再遍历同时缩小遍历的范围;

    代码如下:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <algorithm>
    using namespace std;
    const int MAX = 20010;
    int a[MAX], n, s, ans;
    int main()
    {
        int i, j;
        while (scanf("%d%d", &n, &s) != EOF)
        {
            for (i = 0; i < n; ++i)
            {
                scanf("%d", &a[i]);
            }
            ans = 0;
            sort(a, a + n);
            int temp = n;
            for (i = 0; i < temp; ++i)
            {
                for (j = i + 1; j < temp; ++j)
                {
                    if (a[i] + a[j] <= s)
                    {
                        ans++;
                    }
                    else
                    {
                        temp = j;//缩短遍历的范围
                        break;
                    }
                }
            }
            printf("%d\n", ans);
        }
        return 0;
    }
  • 相关阅读:
    EasyUI DataGrid undefined处理
    EasyUI Tabs
    EasyUI ComboBox ajax
    Spring Controller RequestMapping
    Spring前台填充数据
    关于Map集合的遍历总结
    mvc与三层结构终极区别
    关于cron4j的使用
    中国各个省市县的人口统计,echart展示
    windows 安装nodejs 和 npm
  • 原文地址:https://www.cnblogs.com/lidaojian/p/2486155.html
Copyright © 2011-2022 走看看