LightOJ 1220 Mysterious Bacteria（唯一分解定理 + 素数筛选）

http://lightoj.com/volume_showproblem.php?problem=1220

Mysterious Bacteria

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

Submit Status Practice LightOJ 1220

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

 

 题目大意：

给你一个数x = b^p,求p的最大值

 

x = p1^x1*p2^x2*p3^x3*...*ps^xs

开始我以为是找x1、x2、... 、xs中的最大值，后来发现想错了，x = b^p, x只有一个因子的p次幂构成

如果x = 12 = 2^2*3^1，要让x = b^p,及12应该是12 = 12^1

所以p = gcd(x1, x2, x3, ... , xs);

比如：24 = 2^3*3^1，p应该是gcd(3, 1) = 1,即24 = 24^1

         324 = 3^4*2^2,p应该是gcd(4, 2) = 2,即324 = 18^2

 

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

 

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>

using namespace std;

const int N = 1e5 +10;
const int INF = 0x3f3f3f3f;
typedef long long ll;

int prime[N], k;
bool Isprime[N];

void Prime()
{
    k = 0;
    memset(Isprime, true, sizeof(Isprime));
    prime[1] = false;
    for(int i = 2 ; i < N ; i++)
    {
        if(Isprime[i])
        {
            prime[k++] = i;
            for(int j = i ; 1LL * i * j < N ; j++)
                Isprime[i * j] = false;
        }
    }
}

int gcd(int a, int b)
{
    return a % b == 0 ? b : gcd(b, a % b);
}

int main()
{
    int t, p = 0;
    ll n;//n要用long long 定义，如果n是负数的话会超时
    Prime();
    scanf("%d", &t);
    while(t--)
    {
        p++;
        scanf("%lld", &n);
        int f = 0;

        if(n < 0)
        {
            n = - n;//int定义n这儿会卡住半天出不来，就会超时，为什么这样我也不知道
            f = 1;
        }
        int x, ans = 0;
        for(int i = 0 ; i < k && prime[i] * prime[i] <= n ; i++)
        {
            if(n % prime[i] == 0)
            {
                x = 0;
                while(n % prime[i] == 0)
                {
                    x++;
                    n /= prime[i];
                }
                if(ans == 0)
                    ans = x;
                else
                    ans = gcd(ans, x);
            }
        }
        if(n > 1)
            ans = gcd(ans, 1);
        if(f == 1)
        {
            if(ans % 2 == 0)
               ans = 1;
        }
        printf("Case %d: %d\n", p, ans);
    }
    return 0;
}
/*

8
2147483647
-2147483648
32
-32
64
-64
4
-4

Output:

Case 1: 1
Case 2: 31
Case 3: 5
Case 4: 5
Case 5: 6
Case 6: 3
Case 7: 2
Case 8: 1
*/