UVA 10791 Minimum Sum LCM(质因数分解＋数学分析)

LCM (Least Common Multiple) of a set of integers is defined as the minimum number, which is a multiple of all integers of that set. It is interesting to note that any positive integer can be expressed as the LCM of a set of positive integers. For example 12 can be expressed as the LCM of 1, 12 or 12, 12 or 3, 4 or 4, 6 or 1, 2, 3, 4 etc.

In this problem, you will be given a positive integer N. You have to find out a set of at least two positive integers whose LCM is N. As infinite such sequences are possible, you have to pick the sequence whose summation of elements is minimum. We will be quite happy if you just print the summation of the elements of this set. So, for N = 12, you should print 4+3 = 7 as LCM of 4 and 3 is 12 and 7 is the minimum possible summation.

Input 

The input file contains at most 100 test cases. Each test case consists of a positive integer N ( 1N231 - 1).

Input is terminated by a case where N = 0. This case should not be processed. There can be at most 100 test cases.

Output 

Output of each test case should consist of a line starting with `Case #: ' where # is the test case number. It should be followed by the summation as specified in the problem statement. Look at the output for sample input for details.

Sample Input 

 121050

Sample Output 

 Case 1: 7Case 2: 7Case 3: 6

题意：给定一个数n，求至少两个数的lcm＝n，然后使得它们的和最小。

分析：我们通过lcm的意义可以知道对n进行质因数分解以后，那些数所构成的集合的质因数分解后任意因子的质数已定要等于n质因数分解后相同因子的指数，根据这一条，我们就确定了这些集合中任一因子的指数的最大值，并且根据和最小的思路，那么这个质因数在其他数中已定是0，所以我们就很自然的按照一条思路找到了我们的策略：对n进行质因数分解之后a^q作为单独一个数

////  main.cpp//  Uva 10791////  Created by 张嘉韬 on 16/8/2.//  Copyright © 2016年 张嘉韬. All rights reserved.//#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;typedef unsigned long long ll;ll counter,r;ll init(ll n){    if(n==1) return 2;    counter=0;    r=0;    ll temp=(int)((double)sqrt(n)+1);    for(ll i=2;i<=temp;i++)    {        if(n%i==0)        {            ll ans=1;            counter++;            while(n%i==0)            {                ans=i*ans;                n=n/i;            }            r+=ans;        }    }    if(n!=1) {r+=n;counter++;}    if(counter==1) return r+1;    else return r;}int main(int argc, const char * argv[]) {    //freopen("/Users/zhangjiatao/Documents/暑期训练/input.txt","r",stdin);    ll n;    int t=0;    while(scanf("%lld",&n)==1)    {        //cout<<"shit"<<endl;        if(n==0) break;        t++;        cout<<"Case "<<t<<": "<<init(n)<<endl;    }    return 0;}