1220 - Mysterious Bacteria--LightOj1220 (gcd)

http://lightoj.com/volume_showproblem.php?problem=1220

题目大意： 给你一个x，求出满足 x=b^p, p最大是几。

分析：x=p1^a1*p2^a2*...*pn^an;

p最大是gcd(a1,a2,...,an)。

///他该诉你x,b,p都是整数，所以x,b有可能是负数。当x是负数时，p不能是偶数。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>

using namespace std;
typedef long long LL;
#define N 100000
#define ESP 1e-8
#define INF 0x3f3f3f3f
#define memset(a,b) memset(a,b,sizeof(a))

LL prime
, k;
bool vis
;

void Prime()
{
memset(vis, false);
k = 0;
for(int i=2; i<N; i++)
{
if(vis[i] == 0)
{
prime[k ++] = i;
for(int j=i+i; j<N; j+=i)
{
vis[j] = 1;
}
}
}
}

LL gcd(LL a, LL b)
{
LL r;
while(b)
{
r = a%b;
a = b;
b = r;
}
return a;
}

LL solve(LL n)
{
LL ans, sum;
ans = 0;
sum = 1;
int flag = 0;

for(int i=0; prime[i]*prime[i] <= n; i++)
{
if(n%prime[i] == 0)
{
ans = 0;
while(n%prime[i] == 0)
{
n /= prime[i];
ans ++;
}
if(flag == 0)
{
sum = ans;
flag = 1;
}
else if(flag == 1)
sum = gcd(sum, ans);
}
}
if(n>1)
sum = 1;

return sum;
}

int main()
{
int T, t=1;
Prime();
scanf("%d", &T);
while(T --)
{
LL n;
scanf("%lld", &n);

int b = 0;

if(n<0)
{
n = -n;
b = 1;
}

LL sum = solve(n);

if(b == 1 && sum%2 == 0)
{
while(sum%2 == 0)
sum/=2;
}
printf("Case %d: %lld\n", t++, sum);
}
return 0;
}