PAT(A) - 1029. Median (25)-优快云博客

本文链接：https://blog.youkuaiyun.com/flx413/article/details/52043685

本文介绍了一道算法题目，要求合并两个递增的整数序列并找出中位数。通过使用C++标准库中的向量和排序功能，提供了一个高效的解决方案，并强调了在处理大数据时使用快速输入输出方法的重要性。

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1029. Median (25)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题目要求很简单，给两个整型数序列，将它俩合并，找出中位数。题目测试数据很大，第一次两个点超时了，然后换成scanf 和 printf 格式的输入输出就AC了！所以建议遇到大量数据的时候用C语言的格式化输入输出。

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    //freopen( "123.txt", "r", stdin );
    int n, m, num;
    vector<int> vec;
    int i;
    cin >> n;
    for( i = 0; i < n; i++ ) {
        scanf( "%d", &num );
        vec.push_back( num );
    }

    cin >> m;
    for( i = 0; i < m; i++ ) {
        scanf( "%d", &num );
        vec.push_back( num );
    }

    sort( vec.begin(), vec.end() );

    printf( "%d", vec[(vec.size() - 1)/2] );
    return 0;
}