1029. Median (25)

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
  K接着k个非降序的数
  K2接着k2个非降序的数
  求两队非降序在一起的中间数   如果k+k2为偶数，那么非降序的第(k+k2）/2;否则第（k+k2+1）/2个

评测结果

时间

结果

得分

题目

语言

用时(ms)

内存(kB)

用户

8月01日

17:15

答案正确

​​25​​

​​1029​​

​​C++ (g++ 4.7.2)​​

211

8184

​​datrilla​​

测试点

测试点

结果

用时(ms)

内存(kB)

得分/满分

0

答案正确

1

384

4/4

1

答案正确

1

384

2/2

10

答案正确

210

8032

3/3

11

答案正确

211

8184

1/1

12

答案正确

211

8024

3/3

13

答案正确

1

308

1/1

2

答案正确

1

244

4/4

3

答案正确

1

252

1/1

4

答案正确

1

384

1/1

5

答案正确

1

296

1/1

6

答案正确

102

4176

1/1

7

答案正确

103

4148

1/1

8

答案正确

102

4168

1/1

9

答案正确

103

4088

1/1

#include<iostream>   
using namespace std; 
int main()
{
  int*s1, *s2;
  int len1, len2, index1,index2, now, goal,goalindex; 
  scanf("%d", &len1);
  s1 = (int*)malloc(sizeof(int)*len1);
  for (index1 = 0; index1 < len1; index1++)
     scanf("%d", &s1[index1]);
  scanf("%d", &len2);
  s2 = (int*)malloc(sizeof(int)*len2); 
  for (index2 = 0; index2 < len2; index2++) 
     scanf("%d", &s2[index2]);
  goalindex = (len1 + len2+1) / 2;
  index1 = 0; 
  index2 = 0;
  now = 0;
  while (index1 < len1&&index2 < len2&&now < goalindex)
  { 
    goal = s1[index1] < s2[index2] ? s1[index1++] : s2[index2++];
    now++;
  }
  if (now < goalindex&&index1 < len1)goal = s1[goalindex - now+index1-1];
  else if (now < goalindex&&index2 < len2)goal = s2[goalindex - now + index2-1];
  printf("%d\n",goal);
  free(s2);
  free(s1);
  system("pause");
  return 0;
}