http://lightoj.com/volume_showproblem.php?problem=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 = 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 |
Output for Sample Input |
3 17 1073741824 25 |
Case 1: 1 Case 2: 30
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Case 3: 2 |
PROBLEM SETTER: MUHAMMAD RIFAYAT SAMEE
SPECIAL THANKS: JANE ALAM JAN
任何一个大于1的自然数 ,都可以唯一分解成有限个质数的乘积 ,这里 均为质数,其诸指数 是正整数。
若x是负数,那么所得答案一定要是奇数
这里用ans不断除2得到一个奇数,即答案
#include <bits/stdc++.h>
#define ll long long
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)
{
return b==0?a:gcd(b,a%b);
}
int main()
{
init();
int t,cas=0;
cin>>t;
while(t--)
{
ll n;
int flag=0;
int ans=0;
cin>>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);
}
}