codeforces LCM Challenge
题目:http://codeforces.com/contest/235/problem/A
当给的 n 是奇数时,则是 n * (n - 1) * (n - 2),当是偶数时,则要从 n 往前枚举三个数,求出他们的最小公倍数,感觉最多不会枚举超过10个吧
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 ll gcd(ll a,ll b) 2 { 3 if(!b) return a; 4 else return gcd(b,a % b); 5 } 6 ll lcm(ll a,ll b) 7 { 8 //cout<<"((((\n"; 9 return a / gcd(a,b) * b; 10 } 11 int main() 12 { 13 ll x; 14 ll i,j,k; 15 //freopen("data.txt","r",stdin); 16 while(cin>>x) 17 18 { 19 if(x <= 2) 20 { 21 cout<<x<<endl; 22 continue; 23 } 24 if(x % 2) 25 { 26 cout<<(x) * (x - 1) * (x - 2)<<endl; 27 continue; 28 } 29 ll maxx = -1; 30 for(i = x; i >= x - 10; i--) 31 { 32 for(j = i - 1; j >= x - 10; j--) 33 { 34 for(k = j - 1; k >= x - 10; k--) 35 { 36 if(i < 0 || j < 0 || k < 0) continue; 37 maxx = Max(maxx,lcm(lcm(i,j),k)); 38 } 39 } 40 } 41 cout<<maxx<<endl; 42 43 } 44 return 0; 45 }