Mysterious Bacteria LightOJ - 1220（整数唯一分解定理+筛素数）

最新推荐文章于 2019-01-22 11:39:41 发布

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本文介绍了一种求解特定整数x=b^p中p最大值的算法，通过质因数分解并利用辗转相除法求解最大公约数来找到最大的p值，解决了输入整数为负数等特殊情况。

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Mysterious Bacteria

LightOJ - 1220

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

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

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

本题有好多个坑，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，其次是虽然它给定n<2^32但是不能用int需要用long long才能过，我估计还是负数那里出的问题

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1e5+10;
bool isprime[maxn];
int prime[maxn];
int top;
void init(){
    memset(isprime,true,sizeof(isprime));
    isprime[0] = isprime[1] = false;
    top = 0;
    for(int i = 2; i < maxn; i++){
        if(isprime[i]){
            prime[top++] = i;
            for(int j = i+i; j < maxn; j += i){
                isprime[j] = false;
            }
        }
    }
}
int gcd(int a,int b){
    if(b == 0)
        return a;
    else return gcd(b,a%b);
}
int main(){
    init();
    int t,cas = 0;
    scanf("%d",&t);
    while(t--){
        long long n;
        int flag = 0;
        int ans = 0;
        scanf("%lld",&n);
        if(n < 0){
            flag = 1;
            n = -n;
        }
        for(int i = 0; prime[i] * prime[i] <= n && i < top; i++){
            if(n % prime[i] == 0){
                int cnt = 0;
                while(n % prime[i] == 0){
                    cnt++;
                    n /= prime[i];
                }
                ans = gcd(ans,cnt);
            }
            if(n == 1) break;
        }
        if(n != 1) ans = gcd(ans,1);
        if(flag){
            while(ans % 2 == 0)
                ans /= 2;
        }
        printf("Case %d: %d\n",++cas,ans);
    }
    return 0;
}