Some days ago, I learned the concept of LCM (least common multiple). I‘ve played with it for several times and I want to make a big number with it.
But I also don‘t want to use many numbers, so I‘ll choose three positive integers (they don‘t have to be distinct) which are not greater thann. Can you help me to find the maximum possible least common multiple of these three integers?
9
504
只要这三个数中有两个数是奇数一个是偶数,最小公倍数就是这三个数的积。
#include<stdio.h> int main() { long long LCM,n; while(scanf("%lld",&n)>0) { if(n==1)LCM=1; if(n==2)LCM=2; if(n>2) { if(n%2)LCM=n*(n-1)*(n-2); else { if(n*(n-1)*(n-2)/2<n*(n-1)*(n-3)) LCM=n*(n-1)*(n-3); else LCM=n*(n-1)*(n-2)/2; } } printf("%lld\n",LCM); } }
acd LCM Challenge(求1~n的任意三个数的最大公倍数),布布扣,bubuko.com
acd LCM Challenge(求1~n的任意三个数的最大公倍数)
原文:http://blog.csdn.net/u010372095/article/details/26824135