lightoj 1220 - Mysterious Bacteria（算数基本定理）

最新推荐文章于 2019-07-20 15:11:41 发布

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本文介绍了一种关于细菌RC-01繁殖周期的算法问题，该细菌的生命时长为x天，并能产生p个新细菌，其中x为p的整数次幂。文章通过质因数分解的方法求解x的最大p值，特别考虑了x可能为负数的情况。

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

1073741824

Case 1: 1

Case 2: 30

Case 3: 2

题意：让求x = b^p的最大的p。

这个x是这个细菌能活的天数，而且还能是负的？！有点神奇

就是把x用算数基本定理化成质因数的幂乘积的形式，然后找出所有幂次的最大公约数，就是答案了。

要注意是如果x是负数的话，那么幂次就不能是偶数了，因为一个数的偶数次方肯定是正的嘛，就要把幂次变成奇数再求最大公约数。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long

using namespace std;

const int maxn = 1e6 + 10;

int f[maxn],p[maxn],cnt = 0;

void prime()
{
    LL i,j;
    f[1] = 1;
    for(i=2;i<maxn;i++)
    {
        if(f[i])
            continue;
        p[cnt++] = i;
        for(j=i*i;j<maxn;j+=i)
        {
            f[j] = 1;
        }
    }
}
int gcd(int a,int b)
{
    if(b == 0)
        return a;
    else
        return gcd(b,a%b);
}
int main(void)
{
    int T,i,j;
    LL n;
    prime();
    scanf("%d",&T);
    int cas = 1;
    while(T--)
    {
        scanf("%lld",&n);
        int flag = 0;
        if(n < 0)
        {
            flag = 1;
            n = -n;
        }
        i = 0;
        int g = 0;
        while(p[i]*p[i] <= n && i < cnt)
        {
            int t = 0;
            while(n % p[i] == 0)
            {
                n /= p[i];
                t++;
            }
            i++;
            if(t == 0)
                continue;
            if(flag == 1)
            {
                while(t % 2 == 0)
                {
                    t /= 2;
                }
            }
            g = gcd(g,t);
        }
        if(n > 1)
            g = 1;
        printf("Case %d: %d\n",cas++,g);
    }

    return 0;
}