light_oj 1220 素数分解

light_oj 1220  素数分解

J - Mysterious Bacteria

Time Limit:500MS      Memory Limit:32768KB      64bit IO Format:%lld & %llu

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Description

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

3

17

1073741824

25

Sample Output

Case 1: 1

Case 2: 30

Case 3: 2

题意：给定x，找出最大的p使x=b^p；

思路：素数分解。x=(p1^k1)*(p2^k2)*...*(pn^kn) = (p1^a1*p2^a2*...*pn^an)^p=b^p，即提出来一个最大的p，p=gcd(k1,k2,...,kn)。

注意点：素数分解时由于素数表只能到10^7，所以需要对最后一个因子进行特判。

坑点：x为负数的时候p不能为偶数，这时只要把之前提出来的所有2都乘进去即可，剩下留在外面的奇数p就是答案了。

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#include<string>

#include<math.h>

#include<cctype>



using namespace std;



typedef long long ll;

const int maxn=10001000;

const int INF=(1<<29);

const double EPS=0.0000000001;

const double Pi=acos(-1.0);



ll x,p;

bool isprime[maxn];

vector<ll> prime;



void play_prime()

{

    memset(isprime,1,sizeof(isprime));

    isprime[1]=0;

    for(int i=2;i<maxn;i++){

        if(!isprime[i]) continue;

        for(int j=i+i;j<maxn;j+=i){

            isprime[j]=0;

        }

    }

    for(int i=1;i<maxn;i++){

        if(isprime[i]) prime.push_back(i);

    }

}



int main()

{



    int T;cin>>T;

    play_prime();

    int tag=1;

    while(T--){

        cin>>x;

        bool f=0;

        if(x<0) f=1,x=-x;

        p=-1;

        bool flag=1;

        for(int i=0;i<(int)prime.size()&&prime[i]*prime[i]<=x;i++){

            if(x%prime[i]) continue;

            ll k=0;

            while(x%prime[i]==0){

                x/=prime[i];

                k++;

            }

            if(p==-1) p=k;

            else p=__gcd(p,k);

        }

        if(x!=1) p=1;

        if(f){

            while(p%2==0) p/=2;

        }

        cout<<"Case "<<tag++<<": "<<p<<endl;

    }

    return 0;

}

View Code