UVa OJ - 11321 Sort! Sort!! and Sort!!!

The following program is my ACcepted code for UVA-11321.
It's a for everybody to learn and discuss.
If there is any mistake or comment, please let me know.  :D

此乃UVA 11321 的AC code!

歡迎一同討論學習，如有錯誤與任何建議請留言 : )

點這裡看題目 Click here to see this Problem!

這題跟我之前所解的 UVA 612 DNA Sorting 寫法蠻相似的
請參考這理的解說:

This problem is similar with one of my solved problem:  UVA 612 DNA Sorting
Please see this for reference.


//This program is for UVA 11321 Sort! Sort!! and Sort!!!
//題目來源 Problem link: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=25&page=show_problem&problem=2296

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<vector>
using namespace std;

typedef struct
{
    int num;
    int reminder;
}strSet;

/*
注意: 當你在判斷奇數或偶數時，不要使用 (a.num % 2)==1 來判斷奇數
因為當數字為負數時， (a.num % 2)是 -1
我使用 (a.num % 2)==1 拿了很多次 runtime error (RE)
所以我推薦直接使用 (a.num % 2) 來代表為奇數 (代表1 or -1)
(因為在C++判斷式中，非零就是代表 true!)

Note: when you judge the odd or even, don't use (a.num % 2)==1 to mean the odd!!!
Because when the number could be negative, (a.num % 2) would be -1 !
I used (a.num % 2)==1 to get runtime error (RE) several times.
So I recommend only use (a.num % 2) to mean the odd. (That means 1 or -1)
(In the condition, non-zero represent true!)
*/

bool cmp(strSet a, strSet b)
{
    if(a.reminder != b.reminder)
        return a.reminder < b.reminder;
    if( (a.num % 2) && (b.num % 2) )//both of them are odd
        return a.num > b.num;//larger precede the smaller one
    else if( (a.num % 2 == 0) && (b.num % 2 == 0) )//both of them are even
        return a.num < b.num;//smaller precede the larger one
    else if( (a.num % 2) && (b.num % 2 == 0) )//a is odd, b is even
        return true;//odd number precede the even
    else  if( (a.num % 2 == 0) && (b.num % 2) )//b is odd, a is even
        return false;//odd number precede the even
    return true;
}

int main()
{
    int n, m;
    vector<strSet> set;
   
    while(scanf("%d %d", &n, &m) != EOF)
    {  
        printf("%d %d\n", n, m);
       
        if(n == 0 && m == 0)
            break;
       
        set.clear();
       
        for(int i = 0; i < n; i++)
        {
            strSet tmp;
            scanf("%d", &tmp.num);
            tmp.reminder = tmp.num % m;
            set.push_back(tmp);
        }
       
        //sort set[0] to set[n-1]
        sort(set.begin(), set.end(), cmp);
       
        for(int i = 0; i < n; i++)
            printf("%d\n", set[i].num);
    }
    return 0;
}

Please feel free to use it after adding this blog as an reference. (http://autekroy.blogspot.tw) If there is any mistake or comment, please let me know. :D 

歡迎使用與分享任何內容，但請記得標示此部落格為出處。(http://autekroy.blogspot.tw/) 如果有發現任何的錯誤與建議請留言或跟我連絡。 : )