最小公倍数的最小和（Minimum Sum LCM ） 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <cmath>
 5 using namespace std;
 6 const int maxn = 110;
 7 long long f[maxn], ccount;
 8 
 9 void init(long long n){
10     long long m = (long long)sqrt(n + 0.5);
11     ccount = 0;
12     for (long long i = 2; i <= m && n > 1; i++){
13         if (n%i == 0){
14             long long fac = 1;
15             while (n%i == 0 && n > 1){
16                 fac *= i;
17                 n /= i;
18             }
19             f[ccount++] = fac;    //fac是处理后的因子
20         }
21     }
22     if (n > 1)
23         f[ccount++] = n;
24     /*
25     for (int i = 0; i < ccount; i++){
26         printf("**%lld\n", f[i]);
27     }
28     */
29 }
30 
31 int main(){
32     long long n;
33     int cnt = 0;
34     while (scanf("%lld", &n) && n){
35         cnt++;    
36         init(n);    //将n分解成因子
37         long long ans = 0;
38         if (!ccount || ccount == 1){
39             ans = n + 1;
40         }
41         else{
42             for (int i = 0; i < ccount; i++){
43                 ans += f[i];
44             }
45         }
46         printf("Case %d: %lld\n", cnt, ans);
47     }
48     //system("pause");
49     return 0;
50 }

 

posted @ 2017-10-10 20:51  ouyang_wsgwz  阅读(213)  评论(0)    收藏  举报