LightOJ1220 —— 质因数分解

题目链接：https://vjudge.net/problem/LightOJ-1220

1220 - Mysterious Bacteria

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Time Limit: 0.5 second(s)

Memory Limit: 32 MB

Dr. Mob has just discovered a Deathly Bacteria. He named it RC-01. RC-01 has a very strange reproduction system. RC-01 lives exactly x days. Now RC-01 produces exactly p new deadly Bacteria where x = b^p (where b, p are integers). More generally, x is a perfect p^th power. Given the lifetime x of a mother RC-01 you are to determine the maximum number of new RC-01 which can be produced by the mother RC-01.

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Each case starts with a line containing an integer x. You can assume that x will have magnitude at least 2 and be within the range of a 32 bit signed integer.

Output

For each case, print the case number and the largest integer p such that x is a perfect p^th power.

Sample Input	Output for Sample Input
3 17 1073741824 25	Case 1: 1 Case 2: 30 Case 3: 2

题意：

给出一个数x（可以为负数，绝对值大于等于2）， 求使得满足 x = b^p 的最大p。

题解：

1.对x进行质因子分解。

2.取x所有的质因子个数的最大公约数p，如果x为负数，那么p就只能为奇数，所以一直除以2，直到p为奇数。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e6+;

 bool notprime[MAXN+];
 int prime[MAXN+];
 void getPrime()
 {
     memset(notprime, false, sizeof(notprime));
     notprime[] = notprime[] = true;
     prime[] = ;
     for (int i = ; i<=MAXN; i++)
     {
         if (!notprime[i])prime[++prime[]] = i;
         for (int j = ; j<=prime[ ]&& prime[j]<=MAXN/i; j++)
         {
             notprime[prime[j]*i] = true;
             if (i%prime[j] == ) break;
         }
     }
 }

 int fatCnt;
 LL factor[][];
 void getFactors(LL n)
 {
     LL tmp = n;
     fatCnt = ;
     for(int i = ; prime[i]<=tmp/prime[i]; i++)
     {
         if(tmp%prime[i]==)
         {
             factor[++fatCnt][] = prime[i];
             factor[fatCnt][] = ;
             while(tmp%prime[i]==) tmp /= prime[i], factor[fatCnt][]++;
         }
     }
     if(tmp>) factor[++fatCnt][] = tmp, factor[fatCnt][] = ;
 }

 int gcd(int a, int b)
 {
     return b==?a:gcd(b, a%b);
 }

 int main()
 {
     getPrime();
     int T, kase = ;
     scanf("%d", &T);
     while(T--)
     {
         LL n;
         scanf("%lld", &n);
         getFactors(abs(n));
         LL p = factor[][];
         for(int i = ; i<=fatCnt; i++)
             p = gcd(p, factor[i][]);

         while(n< &&  p%==) p /= ;
         printf("Case %d: %d\n", ++kase, p);
     }
     return ;
 }