c++：百钱百鸡，三层循环到单层循环的解决方法

问题，用100元买100只鸡，公鸡一只5元，母鸡一只3元，小鸡3只一元。问公鸡，母鸡，小鸡各能买多少？请找出所有情况。

如下：

#include<iostream>
using namespace std;
int main()//三层for循环遍历三个对象，即穷举或枚举，直到满足条件的保留，不满足的舍去
{
	int a, b, c;//枚举对象为三个鸡,a为公鸡，b为母鸡，c为小鸡🐥
	for (a = 0; a <= 20; a++)//公鸡枚举的最大范围为20
		for (b = 0; b <= 33; b++)//母鸡为33
			for (c = 0; c <= 100; c++)//小鸡为100
				if (a * 5 + b * 3 + c / 3 == 100 && a + b + c == 100 && c % 3 == 0)//判定条件
					cout << "公鸡只数:"<<a <<'\t' << "母鸡只数:" << b<<'\t' << "小鸡只数:" << c << endl;
	return 0;


}
#include <iostream>
using namespace std;
int main()//双层循环
{
	int a, b, c;
	for (a = 0; a <= 20; a++)
		for (b = 0; b <= 33; b++)
		{
			c = 100 - a - b;
			if (5 * a + 3 * b + c / 3 == 100 && c % 3 == 0)cout << "公鸡只数:" << a << '\t' << "母鸡只数:" << b << '\t' << "小鸡只数:" << c << endl;
		}
	return 0;
}
#include <iostream>//单层循环，数学建模，化简表达式，写出条件
using namespace std;
int main()
{
	int a, b, c;
	for (a = 0; a <= 14; a++)
	{
		b = (100 - 7 * a) / 4;c = 100 - a - b;
		if (b <= 25&&c%3==0&&5*a+3*b+c/3==100)cout << "公鸡只数:" << a << '\t' << "母鸡只数:" << b << '\t' << "小鸡只数:" << c << endl;
		
	}
}