1220 - Mysterious Bacteria(素因子分解)_java素因子分解-优快云博客

本文介绍了一种名为RC-01的虚构致命细菌的繁殖机制，并提供了一个算法来确定这种细菌最大可能的繁殖数量。通过分解给定的生存周期x为质因数并计算这些质因数指数的最大公约数，可以找出最大繁殖次数p。

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1220 - Mysterious Bacteria

PDF (English)

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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

1073741824

Case 1: 1

Case 2: 30

Case 3: 2

题意：给定一个x = b^p；求p的最大值。

思路：n = p1^e1 * p2^e2 * p3^e3 * ..... *pi ^ei

p = gcd(e1，e2， e3，....，ei); 如果n 是负数则需要转换成正数，并且p一定是奇数才可以。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1e7+7;
ll n;
int prim[maxn/10], cnt = 0;
bool vis[maxn];

ll gcd(ll a, ll b) {
    if(!b) return a;
    else return gcd(b, a%b);
}


void getprim(){
    memset(vis, 0, sizeof(vis));
    for(int i = 2; i*i < maxn; i++){
        if(!vis[i]) {
            for(int j = i*i; j < maxn; j += i)
                vis[j] = true;
        }
    }
    for(int i = 2; i < maxn; i++)
        if(!vis[i]) prim[cnt++] = i;
}

ll getpr(ll n) {
    bool f = false;
    if(n < 0) n = -n, f = true;
    ll ans = 0;
    for(int i = 0; i < cnt && n > 1; i++)
    {
        if(n % prim[i] == 0) {
            int c = 0;
            while(n % prim[i] == 0){
                n /= prim[i];
                c++;
            }
            if(ans == 0) ans = c;
            else  ans = gcd(ans, c);
        }
    }
    if(n > 1) ans = 1;
    if(f){
        while(ans % 2 == 0) ans /= 2;
    }
    return ans;
}


int main()
{
    int t, cas = 0;
    getprim();
    cin >> t;
    while(t--){
        cin>>n;
        printf("Case %d: %lld\n", ++cas, getpr(n));
    }
    return 0;
}