最小公倍数的最小和（Minimum Sum LCM ）

复制代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 110;
long long f[maxn], ccount;

void init(long long n){
    long long m = (long long)sqrt(n + 0.5);
    ccount = 0;
    for (long long i = 2; i <= m && n > 1; i++){
        if (n%i == 0){
            long long fac = 1;
            while (n%i == 0 && n > 1){
                fac *= i;
                n /= i;
            }
            f[ccount++] = fac;    //fac是处理后的因子
        }
    }
    if (n > 1)
        f[ccount++] = n;
    /*
    for (int i = 0; i < ccount; i++){
        printf("**%lld\n", f[i]);
    }
    */
}

int main(){
    long long n;
    int cnt = 0;
    while (scanf("%lld", &n) && n){
        cnt++;    
        init(n);    //将n分解成因子
        long long ans = 0;
        if (!ccount || ccount == 1){
            ans = n + 1;
        }
        else{
            for (int i = 0; i < ccount; i++){
                ans += f[i];
            }
        }
        printf("Case %d: %lld\n", cnt, ans);
    }
    //system("pause");
    return 0;
}