题意:
给n个数,求它们的最小公倍数。
分析:
lcm(a,b)==a*b/gcd(a,b);
代码:
//poj 3970//sep9#include <iostream>using namespace std;typedef long long ll;ll gcd(ll a,ll b){return a%b==0?b:gcd(b,a%b);}int main(){int n;while(scanf("%d",&n)==1&&n){ll ans,x;scanf("%lld",&ans);–n;while(n–){scanf("%lld",&x);ans=ans*x/(gcd(ans,x));}if(ans>=1000000) puts("Too much money to pay!");else printf("The CEO must bring %lld pounds.\n",ans);}}
,会得到最大的满足,因为它填补了你的空虚。