九度OJ 1004：Median

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

#define N 1000000

int a1[N+1], a2[N+1];

int cmp(const void *a, const void *b)
{
    return *(int *)a - *(int *)b;
}

int main(void)
{
    int n1, n2, i1, i2, k, id, res;

    while (scanf("%d", &n1) != EOF)
    {
        for(i1=0; i1<n1; i1++)
            scanf("%d", &a1[i1]);
        scanf("%d", &n2);
        for(i2=0; i2<n2; i2++)
            scanf("%d", &a2[i2]);

        qsort(a1, n1, sizeof(a1[0]), cmp);
        qsort(a2, n2, sizeof(a2[0]), cmp);
        a1[n1] = INT_MAX;
        a2[n2] = INT_MAX;

        i1 = i2 = k = 0;
        id = (n1+n2-1)/2;
        while (k < id)
        {
            k ++;
            if (a1[i1] < a2[i2])
                i1 ++;
            else
                i2 ++;
        }
        //printf("i1=%d, i2=%d, k=%d, id=%d\n", i1, i2, k, id);
        if (a1[i1] < a2[i2])
            res = a1[i1];
        else
            res = a2[i2];
        printf("%d\n", res);
    }

    return 0;
}
/**************************************************************
    Problem: 1004
    User: liangrx06
    Language: C
    Result: Accepted
    Time:0 ms
    Memory:8724 kb
****************************************************************/

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：16079

解决：4443

题目描述：

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 non-decreasing 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.

输入：

Each input file may contain more than 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.

输出：

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

样例输入：

4 11 12 13 14
5 9 10 15 16 17

样例输出：

13

来源：

2011年浙江大学计算机及软件工程研究生机试真题

思路：

本题的意思是求两个数组合并后数组的中间数。

实际上不需要全部合并再求，只需要从小到大求出合并后数组，求到中间数时终止即可输出结果。

其实也就是在执行归并排序，该排序有一个小技巧是将两个数组的队尾添加上限数，方便程序书写。

代码：