1029. Median (25)

最新推荐文章于 2020-10-21 17:01:14 发布

dyeDeny

最新推荐文章于 2020-10-21 17:01:14 发布

阅读量497

点赞数

分类专栏： PAT 文章标签： c++ pat

本文链接：https://blog.youkuaiyun.com/dyeDeny/article/details/48231617

版权

PAT 专栏收录该内容

17 篇文章

订阅专栏

本文介绍了一种算法，该算法能够接收两个递增的整数序列，并找出这两个序列合并后的中位数，同时确保合并过程中去除重复元素。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

原题地址：http://www.patest.cn/contests/pat-a-practise/1029

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

解题思路：注意两个序列合并后，重复数字要剔除。

代码如下：

#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <cstring>
#define MAXN 1000005

using namespace std;

int N1,N2;
long int sq[MAXN*2];
long int sq1[MAXN];
long int sq2[MAXN];

/*
int cmp(const void*_a, const void*_b){
    long int a = *(long int *)_a;
    long int b = *(long int *)_b;
    return a > b;   
}
*/

int main(){
    freopen("1029.txt","r",stdin);

    memset(sq,-1,sizeof(sq));
    scanf("%d", &N1);
    for(int i = 0; i < N1; i++) 
        scanf("%ld", &sq1[i]);
   scanf("%d", &N2);
    for(int i = 0; i < N2; i++)
        scanf("%ld", &sq2[i]);

    int p = 0, q = 0;
    int count = 0;
    while(p < N1 || q < N2){
        if(p >= N1){
            if(count == 0) {sq[count++] = sq2[q]; q++;}
            else if(sq[count-1] == sq2[q]) q++;
            else {sq[count++] = sq2[q]; q++; }
        }else if( q >= N2){
            if(count == 0) {sq[count++] = sq1[p]; p++;}
            else if(sq[count-1] == sq1[p]) p++;
            else {sq[count++] = sq1[p]; p++; }
        }else{
        if(sq1[p] < sq2[q]){
            if(count == 0) {sq[count++] = sq1[p]; p++;}
            else if(sq[count-1] == sq1[p]) p++;
            else {sq[count++] = sq1[p]; p++; }
        }else if(sq1[p] > sq2[q]){
            if(count == 0) {sq[count++] = sq2[q]; q++;}
            else if(sq[count-1] == sq2[q]) q++;
            else {sq[count++] = sq2[q]; q++; }
        }else{
            if(count == 0) {sq[count++] = sq2[q]; q++; p++;}
            else if(sq[count-1] == sq2[q]) { q++; p++;}
            else {sq[count++] = sq2[q]; q++; p++; }
        }
        }
    }
//  printf("%d\n",count);
//  for(int i = 0; i<count;i++ ) printf("%ld \n", sq[i]);
    printf("%ld\n",sq[(count*10-5)/20]);
    return 0;
}