light oj 1220

题目连接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=26932

Mysterious Bacteria

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

1073741824

Sample Output

Case 1: 1

Case 2: 30

Case 3: 2

Time Limit: 500MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

题意：给你一个x，x一定能写成x=b^p(b,p都是整数，p当然可以为1)，求最大的p值；

思路：把x因式分解，写成x=prime1^p1+prime2^p2+.....；b是任意的，所以x=b^gcd(p1,p2,p3,p4.........);比如x=2^2 * 3^4 * 5^8，gcd(2,4,8)=2;

所以x=b^2,此时x=(2^1 * 3^2 * 5^4)；这个gcd就是答案p；不过注意题目说x是32int范围内，32位是包括负数的，所以要处理负数，if(x<0) 将x变成正数去

分解因式，最后在答案中去掉所有2，因为负数不可能是一个数的偶数次方，也就是说b^p，p必须是奇数。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<set>
#include<vector>
#include<map>
#include<stack>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=(1<<17)+2333;
const int INF=0x3f3f3f3f;
typedef long long  ll;
int  pri[maxn],vis[maxn];
int a[maxn];
int tot;
void init()
{
    tot=0;
    memset(vis,0,sizeof vis);
    for(int i=2;i<maxn;i++)
    {
        if(!vis[i])
            pri[tot++]=i;
        for(int j=0;j<tot && i*pri[j]<maxn;j++)
        {
            vis[i*pri[j]]=1;
            if(i%pri[j]==0)
                break;
        }
    }
}
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    init();
    int t;
    cin>>t;
    int kase=t;
    while(t--)
    {
        ll  x;
        scanf("%lld",&x);
        int tag=0;
        if(x<0)
            tag=1,x=-x;
        int ans=0;
        for(int i=0;i<tot;i++)
        if(x%pri[i]==0)
        {
            int cnt=0;
            while(x%pri[i]==0)
            {
                cnt++;
                x/=pri[i];
            }
            if(ans==0)
                ans=cnt;
            else ans=gcd(ans,cnt);
            if(x==1)
                break;
        }
        if(x>1)
            ans=1;
        else
        {
            if(tag)
            while(ans%2==0)
            ans/=2;
        }
        printf("Case %d: %d\n",kase-t,ans);
    }
    return 0;
}

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