Bailian 2750: Chicken and Rabbit in the same cage

Total Time Limit: 1000ms Memory Limit: 65536kB

describe

Chickens and rabbits are kept in a cage (chickens have 2 feet, rabbits have 4 feet, no exceptions). Knowing the total number of feet a in the cage, ask how many animals there are at least in the cage and how many animals there are at most.

enter

One line, a positive integer a (a < 32768).

output

A line containing two positive integers, the first is the minimum number of animals, the second is the maximum number of animals, and the two positive integers are separated by a space.
If there is no answer that meets the requirements, output two 0s separated by a space.

sample input

20

Sample output

5 10

Solution: If a is a multiple of 4, obviously there are at least a/4 animals and at most a/2 animals. If a is a multiple of 2 but not a multiple of 4, there are at least a/4+1 animals and at most a/2. 0 0 otherwise.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
using namespace std;

intmain()
{
    int a;
    cin>>a;
    int minx=32768,maxx=0;
    if(a%4==0){
        minx=a/4;
        maxx=a/2;
        printf("%d %d\n",minx,maxx);
    }else if(a%2==0&&a%4!=0){
        minx=a/4+1;
        maxx=a/2;
        printf("%d %d\n",minx,maxx);
    }else{
        printf("0 0\n");
    }
    return 0;
}