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求解方法:
最大公因数(Greatest Common Divisor) 利用“辗转相除法”求,借助递回、非递回思想;
最小公倍数(Least Common Multiple) 通过公式:最大公因数* 最小公倍数=两数乘积 求解。
代码实现
说明:采用C语言,编译环境为DevC++。
#include <stdio.h> #include <stdlib.h> int main() { int m, n, r; int s;
printf("输入两数:"); scanf("%d %d", &m, &n); s = m * n; while(n != 0) { r = m % n;
m = n; n = r; } printf("最大公因数(Greatest Common Divisor):%d\n", m);
printf("最小公倍数(Least Common Multiple):%d\n", s/m); return 0; }
运行结果
写在最后:
读两遍下来,如果仍然有不清楚的地方,可在评论区留言。
如果你有其他感到困惑的问题,欢迎留言。