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1029. Median (25)-PAT甲级真题（two points）

2016-08-16 15:16 501 查看

1029. Median (25)

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

题目大意：给出两个已排序序列，求这两个序列合并后的中间数

分析：用指针pq分别指向两个数组的第一个元素，然后根据中间数就是舍弃掉(n-1)/2个数，所以根据比较大小而向后移动pq指针，表示舍弃掉p和q指针前面的所有数字。当完成后，比较两个序列的指针所指数字的大小，输出较小的那个。如果一个序列已经到达最末端，则输出那个没有到达末端的指针所指向的数字~~~

#include <cstdio>
#include <vector>
using namespace std;
int main() {
int m, n, p = 0, q = 0;
scanf("%d", &m);
vector<long int> v1(m);
for(int i = 0; i < m; i++)
scanf("%ld", &v1[i]);
scanf("%d", &n);
vector<long int> v2(n);
for(int i = 0; i < n; i++)
scanf("%ld", &v2[i]);
int cnt = ((m + n) - 1) / 2;
while(cnt) {
while(p < m && q < n && v1[p] < v2[q] && cnt) {
p++;
cnt--;
}
while(p < m && q < n && v1[p] >= v2[q] && cnt) {
q++;
cnt--;
}
while(p < m && q >= n && cnt) {
p++;
cnt--;
}
while(p >= m && q < n && cnt) {
q++;
cnt--;
}
}
long int ans;
if(p < m && q < n)
ans = v1[p] < v2[q] ? v1[p] : v2[q];
else
ans = p < m ? v1[p] : v2[q];
printf("%ld", ans);
return 0;
}

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