1004_Median

// 1004_Median.cpp : 定义控制台应用程序的入口点。
//题目1004：Median
//时间限制：1 秒内存限制：32 兆特殊判题：否提交：18367解决：5087
//题目描述：
//    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 non-decreasing 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.
//输入：
//    Each input file may contain more than 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.
//输出：
//    For each test case you should output the median of the two given sequences in a line.
//样例输入：
//4 11 12 13 14
//5 9 10 15 16 17
//样例输出：
//13
//来源：
//2011年浙江大学计算机及软件工程研究生机试真题#include "stdafx.h"
#include "stdio.h"#define MAX 1000002long int a[MAX];
long int b[MAX];int main()
{int i,j,count,median;   int n,m;while(scanf("%d",&n)!=EOF){int k=n;int flag = 0;while(k>0)scanf("%ld",&a[n-(k--)]);scanf("%d",&m);k=m;while(k>0)scanf("%ld",&b[m-(k--)]);i=j=count=0;median=(m+n+1)/2;if(!median);else{doif(a[i]>b[j]){j++;count++;if(count == median){printf("%ld\n",b[j-1]);break;}if (j == m){flag = 1;break;}                   }else{i++;count++;if(count == median){printf("%ld\n",a[i-1]);break;}if (i == n){flag = 2;break;}                   }}while(1);if(flag ==1){printf("%ld\n",a[i+median-count]);}else if(flag == 2)printf("%ld\n",b[j+median-count]);}}return 0;
}
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