1029 Median (25 分)

最新推荐文章于 2020-02-12 11:16:50 发布

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最新推荐文章于 2020-02-12 11:16:50 发布

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分类专栏： PAT--*A（感谢柳婼！！！） two points

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博客围绕两个有序数列合并后找中位数展开。介绍了中位数定义，给出输入输出规范，包含示例输入输出。还提及学习了 two points 算法，以及为防止数组越界在数组后加很大的数的技巧。

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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 Specification:

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 (≤2×105) 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 Specification:

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

两个有序数列合并后找中位数（个数为偶数时为左半部分最后一个数）

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

#include<stdio.h>
const int maxn=200005;
const int inf=0x7fffffff;
int s[maxn];///两个数组会内存超限
int main()
{
    int n,m,temp;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
        scanf("%d",&s[i]);
    s[n+1]=inf;///防止越界
    scanf("%d",&m);
    int medianpos=(n+m+1)/2;
    int i=1,count=0;
    for(int j=1; j<=m; j++)
    {
        scanf("%d",&temp);
        while(s[i]<temp)
        {
            count++;
            if(count==medianpos)
                printf("%d",s[i]);
            i++;
        }
        count++;
        if(count==medianpos)
            printf("%d",temp);
    }///第二个数组找完了仍没有找到中位数
    ///就在第一个数组里继续找
    while(i<=n)
    {
        count++;
        if(count==medianpos)
            printf("%d",s[i]);
        i++;
    }
}