ZJU 2124 Perfect Pth Powers

 

Perfect Pth Powers Time limit: 1 Seconds   Memory limit: 32768K  
Total Submit: 2510   Accepted Submit: 488  

We say that x is a perfect square if, for some integer b, x = b2. Similarly, x is a perfect cube if, for some integer b, x = b3. More generally, x is a perfect pth power if, for some integer b, x = bp. 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.

 题目描述：

给出一个int范围的整数x，求x=b^p的最大p值。

分析：

因为x在int范围内，所以p的取值是1到31。枚举一下，用pow函数对x开方，然后验证即可。

注意一点 :该题取值范围若只考虑在int内AC不了。

代码：

#include <stdio.h> 
#include <math.h> 
 
#define int64 long long 
#define I64d "%lld"  
 
int64 POW(int b,int p){  
    int i;  
    int64 s=1;  
    for(i=0;i< p;i++) s*=b;  
    return s;  
}  
 
int main()  
{  
    int64 x,n;  
    int i,y;  
      
    while(scanf(I64d,&n),n){  
        x=n>0?n:-n;  
        for(i=32;i>=2;i--){  
            y=(int)(pow((double)x,1.0/i)+0.5);  
            if(POW(y,i)==x){  
                if(n<0 && i%2==0) continue;  
                else printf("%d/n",i);  
                break;  
            }  
        }  
        if(i==1) printf("1/n");  
    }  
      
    return 0;  
}