唯一分解定理

最新推荐文章于 2025-08-25 21:59:46 发布

原创最新推荐文章于 2025-08-25 21:59:46 发布 · 114 阅读

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该博客探讨了一种数学问题，涉及寻找给定正整数x的最大幂次数p，使得x可以表示为某个整数的p次幂。通过唯一分解定理和最大公约数的计算，解决了负数和正数情况下的问题。文章提供了C++代码实现，并通过一系列测试用例验证了算法的正确性。

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 = bp (where b, p are integers). More generally, x is a perfect pth 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 pth power.

Sample Input

Sample Output

Case 1: 1
Case 2: 30
Case 3: 2

题意：给x,让求满足 x=b^p的p的最大值 （注意x可正可负数）
分析：

x=p1^x1 p2 ^x2 …. pk^xk
b=p1^b1 p2^b2….. pk^bk

分析得：
b1*p=x1 
b2*p=x2 
......
bk*p=xk

p就是(x1,x2,....,xk)的最大公约数.
若为负数,p就必须为奇数,所以先对x化为正数,再对p/2,直到p为奇数
再由唯一分解定理可求

///debug测试样例；
/*
13
17
1073741824
25
2147483647
-2147483648
32
-32
64
-64
4
-4
324
-46656


Case 1: 1
Case 2: 30
Case 3: 2
Case 4: 1
Case 5: 31
Case 6: 5
Case 7: 5
Case 8: 6
Case 9: 3
Case 10: 2
Case 11: 1
Case 12: 2
Case 13: 3
*/

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#define ll long long
using namespace std;
const int maxn=1e6+10;
int vis[maxn],s[maxn],a[maxn];
ll x;
int k=0;

void init()//打表
{
	vis[1]=1;
    for(int i=2; i<=1000000; i++)
        if(vis[i]==0)
        {
            s[k++]=i;
            for(int j=i*2; j<=1000000; j=j+i)
                vis[j]=1;
        }
}
int gcd(int x,int y)//gcd
{
	if(y==0) return x;
	else return gcd(y,x%y);
}
int main()
{
	int t,op=1;
	scanf("%d",&t);
	init();
	while(t--)
	{
		scanf("%lld",&x);
		int flag=0;
		if(x<0)//负数时
		{
			x=-x;
			flag=1;
		}
		int ans=0;//p的值
		for(int i=0;s[i]*s[i]<=x&&i<k;i++)
		{
			int res=0,f=0;
			while(x%s[i]==0)
			{
				f=1;res++;
				x/=s[i];
			}
			if(f)
			{
				if(ans==0) ans=res;
				else ans=gcd(res,ans);
			}
			if(x==0||x==1) break;
		}
		if(x>1) //因子是它本身时
			ans=1;
		if(flag)//负数
			while(ans%2==0)
				ans/=2;
		printf("Case %d: %d\n",op++,ans);
	}
}