最小公倍数的最小和（Minimum Sum LCM ）

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 using namespace std;
 const int maxn = ;
 long long f[maxn], ccount;

 void init(long long n){
     long long m = (long long)sqrt(n + 0.5);
     ccount = ;
     for (long long i = ; i <= m && n > ; i++){
         if (n%i == ){
             long long fac = ;
             while (n%i ==  && n > ){
                 fac *= i;
                 n /= i;
             }
             f[ccount++] = fac;    //fac是处理后的因子
         }
     }
     if (n > )
         f[ccount++] = n;
     /*
     for (int i = 0; i < ccount; i++){
         printf("**%lld\n", f[i]);
     }
     */
 }

 int main(){
     long long n;
     int cnt = ;
     while (scanf("%lld", &n) && n){
         cnt++;
         init(n);    //将n分解成因子
         long long ans = ;
         if (!ccount || ccount == ){
             ans = n + ;
         }
         else{
             for (int i = ; i < ccount; i++){
                 ans += f[i];
             }
         }
         printf("Case %d: %lld\n", cnt, ans);
     }
     //system("pause");
     return ;
 }