Openjudge noi 2.1 1752: chicken and rabbit in the same cage

【 Topic link 】

OpenJudge NOI 2.1 1752: Chicken and rabbit in the same cage

【 Topic test site 】

1. enumeration

【 Their thinking 】

solution 1： enumeration

With chicken x only , The rabbit y only , It is known that there is a One foot , So there are
$2x+4y=a$
Chicken has at least 0 only , When all feet are chicken feet , Most chickens , Yes $\frac{a}{2}$ only , therefore $0\le x\le \frac{a}{2}$
Rabbits have at least 0 only , At most $\lfloor \frac{a}{4}\rfloor$ only , therefore $0\le y\le \lfloor \frac{a}{4}\rfloor$
Summed up in ：

• enumerable object ：x, y
• Enumeration range ：$0\le x\le \frac{a}{2}$,$0\le y\le \lfloor \frac{a}{4}\rfloor$
• Enumeration condition ：$2x+4y=a$
  Select all that meet the conditions x And y in ,x+y Maximum and minimum values of .

solution 2： Looking for a regular

If the number of input feet is odd , Then there is no answer that meets the requirements .
When there are as many chickens as possible , Animals should be the most . When there are as many rabbits as possible , Animals should be the least .
Chicken up to $\frac{a}{2}$, Then the animals are at most $\frac{a}{2}$
Rabbits are at most $\lfloor \frac{a}{4}\rfloor$, be left over $a\%4$ Every foot is a chicken , Yes $\frac{a\%4}{2}$ only , At this time, the animals are at least , Yes $\lfloor \frac{a}{4}\rfloor +\frac{a\%4}{2}$ only .

【 Solution code 】

solution 1： enumeration

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
int main()
{
    
    int a, mx = 0, mn = INF;
    cin >> a;
    for(int x = 0; x <= a/2; ++x)
        for(int y = 0; y <= a/4; ++y)
        {
    
            if(2*x+4*y == a)
            {
    // Update the maximum and minimum values 
                mx = max(mx, x+y); 
                mn = min(mn, x+y);
            }
        }
    if(mx == 0 && mn == INF)// If not found  
        cout << "0 0";
    else
        cout << mn << ' ' << mx;
    return 0;
}

solution 2： Looking for a regular

#include<bits/stdc++.h>
using namespace std;
int main()
{
    
    int a;
    cin >> a;
    if(a % 2 == 1) 
        cout << "0 0";
    else
        cout << a/4+a%4/2 << ' ' << a/2;
    return 0;
}