LightOJ1220 —— 质因数分解

题目链接：https://vjudge.net/problem/LightOJ-1220

1220 - Mysterious Bacteria

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Time Limit: 0.5 second(s)

Memory Limit: 32 MB

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
3 17 1073741824 25	Case 1: 1 Case 2: 30 Case 3: 2

题意：

给出一个数x（可以为负数，绝对值大于等于2）， 求使得满足 x = b^p 的最大p。 

题解：

1.对x进行质因子分解。

2.取x所有的质因子个数的最大公约数p，如果x为负数，那么p就只能为奇数，所以一直除以2，直到p为奇数。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 1e6+10;

bool notprime[MAXN+1];
int prime[MAXN+1];
void getPrime()
{
    memset(notprime, false, sizeof(notprime));
    notprime[0] = notprime[1] = true;
    prime[0] = 0;
    for (int i = 2; i<=MAXN; i++)
    {
        if (!notprime[i])prime[++prime[0]] = i;
        for (int j = 1; j<=prime[0 ]&& prime[j]<=MAXN/i; j++)
        {
            notprime[prime[j]*i] = true;
            if (i%prime[j] == 0) break;
        }
    }
}

int fatCnt;
LL factor[1000][2];
void getFactors(LL n)
{
    LL tmp = n;
    fatCnt = 0;
    for(int i = 1; prime[i]<=tmp/prime[i]; i++)
    {
        if(tmp%prime[i]==0)
        {
            factor[++fatCnt][0] = prime[i];
            factor[fatCnt][1] = 0;
            while(tmp%prime[i]==0) tmp /= prime[i], factor[fatCnt][1]++;
        }
    }
    if(tmp>1) factor[++fatCnt][0] = tmp, factor[fatCnt][1] = 1;
}

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

int main()
{
    getPrime();
    int T, kase = 0;
    scanf("%d", &T);
    while(T--)
    {
        LL n;
        scanf("%lld", &n);
        getFactors(abs(n));
        LL p = factor[1][1];
        for(int i = 2; i<=fatCnt; i++)
            p = gcd(p, factor[i][1]);

        while(n<0 &&  p%2==0) p /= 2;
        printf("Case %d: %d
", ++kase, p);
    }
    return 0;
}