UVA 10622 Perfect P-th Powers （唯一分解定理 + GCD）

题意：找所给定的一个数变成x的p次方的形式， 寻找P最大是多少。

解题思路：先使用唯一分解定理分解素因子， 然后再求所有素因子在一起的GCD，注意负数的P不能为偶数只能为奇数。边分解边求GCD， 否则可能超时。

We say that x is a perfect square if, for some integer b, x = b
2
.
Similarly, x is a perfect cube if, for some integer b, x = b
3
.
More generally, x is a perfect pth power if, for some integer b,
x = b
p
. Given an integer x you are to determine the largest
p such that x is a perfect pth power.
Input
Each test case is given by a line of input containing x. The
value of x will have magnitude at least 2 and be within the
range of a (32-bit) int in C, C++, and Java. A line containing
`0' follows the last test case.
Output
For each test case, output a line giving the largest integer p
such that x is a perfect pth power.
Sample Input
17
1073741824
25
0
Sample Output
1
30
2

#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<cctype>#include<list>#include<iostream>#include<map>#include<queue>#include<set>#include<stack>#include<vector>using namespace std;#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)int buf[10];inline long long read(){    long long x=0,f=1;    char ch=getchar();    while(ch<'0'||ch>'9')    {        if(ch=='-')f=-1;        ch=getchar();    }    while(ch>='0'&&ch<='9')    {        x=x*10+ch-'0';        ch=getchar();    }    return x*f;}inline void writenum(int i){    int p = 0;    if(i == 0) p++;    else while(i)        {            buf[p++] = i % 10;            i /= 10;        }    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);}/**************************************************************/#define MAX_N 1000000const int INF = 0x3f3f3f3f;int isprime[MAX_N];int prime[MAX_N];int cnt = 0;int flag = 0;void make_prime(){    cnt = 0;    memset(isprime, 1, sizeof(isprime));    isprime[1] = 0;    for(long long i = 2 ; i < MAX_N ; i++)    {        if(isprime[i])        {            prime[cnt++] = i;            for(long long j = i * i ; j < MAX_N; j += i)            {                isprime[j] = 0;            }        }    }}long long gcd(long long a, long long b){    return b ? gcd(b, a % b) : a;}long long f(long long n){    long long ret = 0;    for(int i = 0 ; i < cnt && prime[i] <= n ; i++)    {        long long cnt = 0;        while(n % prime[i] == 0)        {            cnt++;            n /= prime[i];        }        ret = gcd(cnt, ret);    }    if(ret == 0) return 1;    if(flag)    {        while(ret % 2 == 0)        {            ret /= 2;        }    }    return ret;}int main(){//    freopen("in.txt", "r", stdin);//    freopen("out1.txt", "w", stdout);    make_prime();    long long n;    while(~scanf("%lld", &n) && n)    {        flag = 0;        if(n < 0)        {            flag = 1;            n = -n;        }        long long ans = f(n);        printf("%lld\n", ans);    }    return 0;}