A1029. Median

本文介绍了一种高效算法,用于找出两个已排序整数序列的中位数。通过合并序列并逐步比较元素,该算法能在较短时间内找到正确答案。

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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

代码:

#include <cstdio>

const int maxn = 1000010;
const int INF = 0x7fffffff;  // int数据上限

int main() {
    int n, m;
    int s1[maxn], s2[maxn];
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &s1[i]);
    }
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
        scanf("%d", &s2[i]);
    }
    s1[n] = s2[m] = INF;  // 两个序列的最后一个元素设为int上限
    int medianPos = (n + m - 1) / 2;
    int i = 0, j = 0, count = 0;
    while (count < medianPos) {
        if (s1[i] < s2[j]) {
            i++;
        } else {
            j++;
        }
        count++;
    }
    if (s1[i] < s2[j]) {
        printf("%d\n", s1[i]);
    } else {
        printf("%d\n", s2[j]);
    }
    return 0;
}
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