奇偶排序

输入若干非负整数数字，请先取出奇数数字按从大到小排序，再取出偶数数字，从小到大排序。

样例输入：12 34 5 7 92 3 8

样例输出： 7 5 3 8 12 34 92

 

不同语言的解法

1，使用C语言来解决

#include <stdio.h>

#include <stdlib.h>

#include <string.h>



// little -> big

int comp1(const void *a, const void *b){

	return *(int *)a - *(int *)b;

}

// big -> little

int comp2(const void *a, const void *b){

	return *(int *)b - *(int *)a;

}

int main(){

	int odd[1000] = {0};

	int even[1000] = {1};



	int input = 0;



	int odd_idx = 0;

	int even_idx = 0;



	int i = 0;

	char a;



	do{	

		scanf("%d",&input);

		if(input % 2){

			odd[odd_idx] = input;

			odd_idx++;

		}else{

			even[even_idx] = input;

			even_idx++;

		}

		a = getchar();

		if(a == '\n')

			break;

	}while(1);



	qsort(odd, odd_idx, sizeof(int), comp2);

	qsort(even, even_idx, sizeof(int), comp1);



	for(i=0; i<odd_idx; i++){

		printf("%d ",odd[i]);

	}



	for(i=0; i<even_idx; i++){

		printf("%d ", even[i]);

	}



	return 0;

}

 2， 使用C++来完成

#include <iostream>

#include <algorithm>



using namespace std;



int main(){

	int input;

	int odd[1000] = {1};

	int even[1000] = {0};

	

	int odd_idx = 0;

	int even_idx = 0;

	

	int i = 0;

	

	while(cin>>input, cin.get() != '\n'){

		if(input % 2){

			odd[odd_idx++] = input;

		}else{

			even[even_idx++] = input;

		}

		if(input == '\n'){

			break;

		}

	}

	

	sort(odd, odd + odd_idx);

	sort(even, even + even_idx);

	

	for(i=0; i<odd_idx; i++){

		cout<<odd[i]<<" ";

	}

	for(i=0; i<even_idx; i++){

		cout<<even[i]<<" ";

	}



	return 0;

}

 使用Python来完成

import sys

line = sys.stdin.readline()

num = [int(x) for x in line.split(' ') if line.strip()]

odd = [ x for x in num if x%2 == 1]

even = [x for x in num if x%2 == 0]

odd.sort(reverse=True)

even.sort()

print(odd,even)