鸡兔同笼问题 C语言解法

#include <stdio.h>

#pragma warning(disable:4996)

int xy(int a1, int a2, int b1, int b2, int c1, int c2);

int main() { int a1 = 4, b1 = 2; // 兔子四条腿，鸡两条腿 int a2 = 1, b2 = 1; // 兔子和鸡的数量系数 int c1, c2;

printf("鸡兔同笼问题：\n");
printf("请输入兔子和鸡的总数：");
scanf("%d", &c2);
printf("请输入兔子的腿和鸡脚数的总和：");
scanf("%d", &c1);

int x = xy(a1, a2, b1, b2, c1, c2); // 计算兔子数量
int y = c2 - x; // 计算鸡数量

printf("兔子共有%d只，鸡共有%d只。\n", x, y);

return 0;

}

int xy(int a1, int a2, int b1, int b2, int c1, int c2) { int a3 = a1 * b2, a4 = a2 * b1; int b3 = b1 * b2, b4 = b2 * b1; int c3 = c1 * b2, c4 = c2 * b1;

return (c3 - c4) / (a3 - a4); // 返回兔子数量

}