【九度OJ】1004：Median

地址：
http://ac.jobdu.com/problem.php?pid=1004
題目描述：
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.
輸入：
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.
輸出：
For each test case you should output the median of the two given sequences in a line.
樣例輸入：
4 11 12 13 14
5 9 10 15 16 17
樣例輸出：
13
來源：
2011年浙江大學計算機及軟件工程研究生機試真題

源碼：

#include<stdio.h>

int na, nb;
int median;
int strA[ 1000005 ];
int strB[ 1000005 ];
int strAB[ 1000005 ];

int main(){
    while( scanf( "%d", &na) != EOF ){
        for( int i = 0; i < na; i ++ ){
            scanf( "%d", &strA[ i ] );
        }
        scanf( "%d", &nb);
        for( int i = 0; i < nb; i ++ ){
            scanf( "%d", &strB[ i ] );
        }

        int median = (na + nb)/2;

        for( int i = 0, x = 0, y = 0; i < median + 1; i ++ ){
            if( x > na - 1 ){
                strAB[ i ] = strB[ y ];
                y++;
                continue;
            }
            if( y > nb - 1 ){
                strAB[ i ] = strA[ x ];
                x++;
                continue;
            }
            if( strA[ x ] > strB[ y ] ){
                strAB[ i ] = strB[ y ];
                y++;
            }
            else{
                strAB[ i ] = strA[ x ];
                x++;
            }
        }

        if( (na+nb) % 2 == 1 ){   //奇數
            printf("%d\n", strAB[ median ] );
        }
        else{   //偶數
            printf("%d\n", strAB[ median - 1 ] );
        }

    }
}
/**************************************************************
    Problem: 1004
    User: 螺小旋
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:12740 kb
****************************************************************/