LightOJ-1220 Mysterious Bacteria （素数打表+欧几里得算法+唯一分解定理）给出x,求x=a^p,最大的指数

题目大意：

x = b^p, x只有一个因子的p次幂构成
如果24 = 2^33^1，p应该是gcd(3, 1) = 1,即24 = 24^1
324 = 3^42^2=(3^22)^2,p应该是gcd(4, 2) = 2,即324 = 18^2
所以p = gcd(x1, x2, x3, ... , xn){欧几里得算法求取最大公约数};
本题有一个坑，就是x可能为负数，如果x为负数的话，x = b^q, q必须使奇数，所以将x转化为正数求得的解如果是偶数的话必须将其一直除2转化为奇数

题目：

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 xdays. Now RC-01 produces exactly p new deadly Bacteria where x = bp (where b, p are integers). More generally, x is a perfect pth 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 pth power.

Sample Input

1073741824

Sample Output

Case 1: 1

Case 2: 30

Case 3: 2

AC代码：

#include<iostream>
#include<string.h>
typedef long long ll;
#include<stdio.h>
using namespace std;
#define M 1000010
int dp[M];
int book[M];
int t,k;
ll m;
void dfs()
{k=0;memset(dp,0,sizeof(dp));memset(book,0,sizeof(book));for(int i=2; i<M; i++)/**因为任何一个整数都可以由一个素数经过乘法运算得到，故可通过素数打表的方式求得某数的唯一分解（得到幂次最大）*/if(!book[i]){dp[k++]=i;/*记录素数*/for(int j=2; i*j<M; j++)book[j*i]=1;}
}
int gcd(int a,int b)
{return !b?a:gcd(b,a%b);
}
int main()
{cin>>t;int tt=1;dfs();while(t--){int flag=0;cin>>m;if(m<0)/**本题的坑：m可能为负数*/{flag=1;m=-m;}int ans=0;for(int i=0;i<k&&dp[i]*dp[i]<=m;i++)/**care:remember停止条件为i<k&&dp[i]*dp[i]<=m，缺一不可*/{if(m%dp[i]==0){int a=0;while(m%dp[i]==0){m/=dp[i];/**唯一分解定理：直接对m的值进行操作，得到x=a1^b1*a2^b2.....an^bn*/a++;}if(ans=0)ans=a;elseans=gcd(ans,a);/**欧几里得算法：求得到的素数的幂次进行求取最大公约数操作*/}if(m==1)break;}if(m>1)/**则m为素数，因数只有1和自身*/ans=1;if(flag)/**填坑：若为负数，幂次不可能为偶数，故由最大公约数求出最大的奇数*/{while(ans%2==0)ans/=2;}printf("Case %d: %d\n",tt++,ans);}return 0;
}