zoukankan      html  css  js  c++  java
  • 1029. Median (25)

    Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.

    Given two increasing sequences of integers, you are asked to find their median.

    Input

    Each input file contains one test case. Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (<=1000000) is the size of that sequence. Then N integers follow, separated by a space. It is guaranteed that all the integers are in the range of long int.

    Output

    For each test case you should output the median of the two given sequences in a line.

    Sample Input
    4 11 12 13 14
    5 9 10 15 16 17
    
    Sample Output
    13
    

    排序
    代码:
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <set>
    #include <algorithm>
    using namespace std;
    
    long long s[2000000];
    int main()
    {
        int a,b;
        scanf("%d",&a);
        for(int i = 0;i < a;i ++)
        {
            scanf("%lld",&s[i]);
        }
        scanf("%d",&b);
        b += a;
        for(int i = a;i < b;i ++)
        {
            scanf("%lld",&s[i]);
        }
        sort(s,s+b);
        printf("%lld",s[(b-1)/2]);
    }

     维护大小为(a+b)/2的堆。

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <queue>
    #include <algorithm>
    using namespace std;
    int main() {
        priority_queue<int,vector<int>,less<int> >q;
        int a,b,d,mid;
        scanf("%d",&a);
        for(int i = 0;i < a;i ++) {
            scanf("%d",&d);
            q.push(d);
        }
        scanf("%d",&b);
        mid = a + b;
        mid = (mid % 2) ? (mid / 2 + 1) : (mid / 2);
        while(q.size() > mid)q.pop();
        while(b && q.size() < mid) {
            scanf("%d",&d);
            b --;
            q.push(d);
        }
        for(int i = 0;i < b;i ++) {
            scanf("%d",&d);
            if(q.top() > d) {
                q.pop();
                q.push(d);
            }
        }
        printf("%d",q.top());
    }
  • 相关阅读:
    hdu 3336 Count the string KMP+DP优化
    Codeforces Round #345 (Div. 1) A. Watchmen 模拟加点
    Codeforces Round #345 (Div. 1) B. Image Preview
    大数据时代下EDM邮件营销的变革
    如何选择EDM电子邮件服务提供商
    一般邮件营销平台可以获取的三个参数
    EDM邮件营销真的落伍了吗?
    EDM概念之A/B分类测试法
    EDM营销技巧之如何进行用户唤醒
    如何优化电子邮件营销的效果
  • 原文地址:https://www.cnblogs.com/8023spz/p/7967799.html
Copyright © 2011-2022 走看看