Electronic Association C language level 2 60, integer parity sort (real question in June 2021)

/*
 Electronics Association  C Language  2 level   60 、 Integer parity sorting （2021 year  6 The real question of the month ）

 Given  10 A sequence of integers , Ask to reorder it . Sorting requirements :
1. Odd number first , Even numbers come after ;
2. Odd numbers are sorted in descending order ;
3. Even numbers are sorted in reverse order of input .
 Input ： The input line , contain  10  It's an integer , Separate each other with a space ,
 The range of each integer is greater than or equal to  0,
 Less than or equal to  100.
 Output ： After sorting as required, a row containing sorted  10  It's an integer ,
 Numbers are separated by a space .
 The sample input 
4 7 3 13 11 12 0 47 34 98
 Sample output 
47 13 11 7 3 98 34 0 12 4
*/
#include<iostream>
#include<algorithm>
using namespace std;
bool cmp(int a,int b)
{
	return a>b;
}
int main() 
{
	int a[10+10],b[10+10];
	int m=0,n=0,temp,i;
	for(i=1; i<=10; i++)
	{	
		cin>>temp;
		if(temp%2==1)
		{ 
			a[m]=temp;
			m++; 
		}
		else
		{ 
			b[n]=temp;n++; 
		}
	}
	
	sort(a,a+m,cmp); // Array elements are sorted from large to small 
	for(i=0; i<m; i++)
		cout<<a[i]<<" ";
	
	for(i=0; i<n; i++)
		cout<<b[n-1-i]<<" "; // Sort in reverse order according to the input order   Reverse output 
	
	return 0;
}
/*
 Homework ：
1.10 Simple sorting of programming basics  06 Integer parity sorting 
http://noi.openjudge.cn/ch0110/06/
http://bailian.openjudge.cn/practice/3682/
http://bailian.openjudge.cn/practice/2871/ 
*/

---

---

1.10 Simple sorting of programming basics 06 Integer parity sorting

OpenJudge - 06: Integer parity sorting

OpenJudge - 3682: Integer parity sorting

---

C++ Code ： Method 1

/* 
1.10 Simple sorting of programming basics  06 Integer parity sorting   Method 1 
http://noi.openjudge.cn/ch0110/06/
http://bailian.openjudge.cn/practice/3682/
http://bailian.openjudge.cn/practice/2871/ 
*/
#include <bits/stdc++.h>
using namespace std;

int main() 
{
	int a[100], b[100];
	for (int i = 0; i < 10; i++) 
	{
		cin >> a[i];
 	}
	int cnt=0;
	for(int i=0;i<10;i++){
		if(a[i]%2==1)
		{
 			b[cnt]=a[i];
 			cnt++;
 		}
 	}
 	sort(b,b+cnt,greater<int>());
	int temp=cnt;
	for(int i=0;i<10;i++)
	{
		if(a[i]%2==0)
		{
 			b[cnt]=a[i];
 			cnt++;
 		}
 	}
	
	sort(b+temp,b+10);
	
	for (int i = 0; i < 10; i++) 
	{
 		cout << b[i] << " ";
 	}
 	cout << endl;
 	
	 return 0;
}

---

---

C++ Code ： Method 2

/*
1.10 Simple sorting of programming basics _06 Integer parity sorting   Method 2 
http://noi.openjudge.cn/ch0110/06/
 Total time limit : 1000ms  Memory limit : 65536kB
 describe 
 Given 10 A sequence of integers , Ask to reorder it . Sorting requirements :

1. Odd number first , Even numbers come after ;

2. Odd numbers are sorted in descending order ;

3. Even numbers are sorted in descending order .

 Input 
 The input line , contain 10 It's an integer , Separate each other with a space , The range of each integer is greater than or equal to 0, Less than or equal to 100.
 Output 
 Sort as required and output a line , Include sorted 10 It's an integer , Numbers are separated by a space .
 The sample input 
4 7 3 13 11 12 0 47 34 98
 Sample output 
47 13 11 7 3 0 4 12 34 98
 source 
1873
*/
#include <iostream>
#include <cstdio>
using namespace std;
int a[105];
int b[105];
int main()
{
    int c,i,j=0,k=0;
	for(int i=0; i<10; i++)
	{
	    cin >> c;
		if(c%2==1)
		    a[j++]=c;
		else
		    b[k++]=c;
	}
	
	for(int i=0; i<j; i++)
	{
		for(int l=0; l<j-1-i; l++)
		{
			if(a[l]<a[l+1])
			{
			   swap(a[l],a[l+1]);
		    }
		}
	}
	
	for(int i=0; i<k; i++)
	{
		for(int l=0; l<k-1-i; l++)
		{
			if(b[l]>b[l+1])
			{
			   swap(b[l],b[l+1]);
		    }
		}
	}
	
	for(int i=0; i<j; i++)
	{
		cout << a[i] << " ";
	}
	
	for(int i=0; i<k; i++)
	{
		cout << b[i] << " ";
	}
	
	return 0;
		
} 

---

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python3 Code ：

"""
1.10  Simple sorting of programming basics  06  Integer parity sorting 
http://noi.openjudge.cn/ch0110/06/

"""
a=list(map(int,input().split()))
 
lena=len(a)

ans01=[]
ans02=[]

i=0
while i<lena:
            if a[i]%2==0:
                        ans01.extend([int(a[i])])
                        
            else:
                        ans02.extend([int(a[i])])
            i+=1
ans01.sort()
ans02.sort()
#print(ans01)
#print(ans02)

j=len(ans02)-1
while j>=0:
            print("%d" %ans02[j],end=" ")
            j-=1
k=0
while k<len(ans01):
            print("%d" %ans01[k],end=" ")
            k+=1

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