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)