鸡兔同笼问题求解 C语言代码

#include <stdio.h>

#pragma warning(disable:4996)

int xy(); int main() { int x, y, a1, a2, b1, b2, c1, c2; int a3, b3, c3; int a4, b4, c4;

a1 = 4; b1 = 2;
a2 = 1; b2 = 1;

printf('鸡兔同笼问题：');
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);

xy(a1, a2, a3, a4, b1, b2, c1, c2);

if (x >= 0 && y >= 0 && (x + y) == c2 && (a1 * x + b1 * y) == c1) {
    printf('   共有%d只鸡，%d只兔子。\n', y, x);
} else {
    printf('   无解！\n');
}

return 0;

}

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

a3 = a1 * a2; a4 = a2 * a1;
b3 = b1 * a2; b4 = b2 * a1;
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);
return 0;

}