鸡兔同笼问题 C语言解法

#include <stdio.h>

#pragma warning(disable:4996)

void calculate(int a1, int a2, int b1, int b2, int c1, int c2, int* x, int* y);

int main() { int a1 = 4, a2 = 2, b1 = 1, b2 = 1, 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, y;
calculate(a1, a2, b1, b2, c1, c2, &x, &y);

printf('ikun共有%d只，兔子共有%d只。', y, x);

return 0;

}

void calculate(int a1, int a2, int b1, int b2, int c1, int c2, int* x, int* y) { int a3 = a1 * a2, a4 = a2 * a1; int b3 = b1 * a2, b4 = b2 * a1; int c3 = c1 * a2, c4 = c2 * a1; *y = (c3 - c4) / (b3 - b4);

a3 = a1 * b2; a4 = a2 * b1;
b3 = b1 * b2; b4 = b2 * b1;
c3 = c1 * b2; c4 = c2 * b1;
*x = (c3 - c4) / (a3 - a4);

}