1029. Median (25)

题目链接：http://www.patest.cn/contests/pat-a-practise/1029

题目：

1029. Median (25)

时间限制

400 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

13

分析：

找出两个数组的共同中位数。能够用递归来做：先找出各自数组的中位数，然后比較，依据结果的大小（都是相异的数字，假设同样则直接就是中位数）来确定保留左边还是右边这段（当然原数组是排好序的）。再接着对子串进行递归查找。

演示样例：

比方题目中的Sample。对于11 12 13 14。它的中位数是12，第二个数组9 10 15 16 17，它的中位数是15，由于15大于12，所曾经者取后半部分13 14，后者取前半部分9 10 15，继续，前者中位数是13，后者中位数是10，13大于10，所曾经者取前半部分13。后则会取后半部分10 15，由于前者仅仅有一个数字13，把它和后者的剩余部分组成一个新的数组：10 13 15（排好序）。中位数就能够用下标直接找到为13。

AC代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
long nums1[1000001];
long nums2[1000001];
long all[1000001];
bool cmp (long A, long B){
 return A < B;
}
int start, End, mid;
long *a;
long *b;
void find_Med(int s1, int e1, int s2, int e2){
 int m1 = (s1 + e1) / 2;
 int m2 = (s2 + e2) / 2;
 if (s1 == e1){//假设短数组仅仅有一个数字，则记录其和长数组的两端下标返回
  mid = s1;
  start = s2;
  End = e2;
  return;
 }
 else if (a[m1] == b[m2]){//假设遇到两个数字同样，则其就是中位数
  mid = m1;
  start = m2;
  End = m2;
  return;
 }
 else if (a[m1] < b[m2]){//假设数组1的中位数小于数组2的中位数。取数组1的后半部分和数组2的前半部分继续查找
  find_Med(m1+ 1, e1, s2, e2 - m1 + s1 - 1);
 }
 else if (a[m1] > b[m2]){//假设数组1的中位数大于数组2的中位数，取数组1的前半部分和数组2的后半部分继续查找
  find_Med(s1, m1, s2 + e1 - m1, e2);
 }
}
int main(void){
 //freopen("F://Temp/input.txt", "r", stdin);
 int n1, n2;
 while (scanf("%d", &n1) != EOF){
  for (int i = 0; i < n1; i++){
   scanf("%ld", &nums1[i]);
  }
  scanf("%d", &n2);
  for (int i = 0; i < n2; i++){
   scanf("%ld", &nums2[i]);
  }
  int n_max, n_min;
  if (n1 <= n2){
   a = nums1;
   b = nums2;
   n_min = n1;
   n_max = n2;
  }
  else{
   a = nums2;
   b = nums1;
   n_min = n2;
   n_max = n1;
  }
  find_Med(0, n_min - 1, 0, n_max - 1);
  int i;
  for (i = 0; i < End - start + 1; i++){
   all[i] = b[start + i];
  }
  all[i] = a[mid];//把短数组的最后一个数插入到长数组的最后剩余一部分中
  sort(all, all + i + 1,cmp);//新组成的尾巴数组排序
  printf("%ld
", all[i / 2]);
 }
 return 0;
}

截图：

——Apie陈小旭