[暑假集训--数论]poj1730 Perfect Pth Powers

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 p th 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 p th power.

Sample Input

Sample Output

1
30
2

给个n，把它表示成a^b的形式，问b最大是多少，n有负数

正数简单，负数要打个标记，最后答案b要除到没有2的因子为止

比如-1073741824=-(2^30)=(-4)^15，但是不能是(-2)^30

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 #define LL long long
 5 inline LL read()
 6 {
 7     LL x=0,f=1;char ch=getchar();
 8     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 9     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
10     return x*f;
11 }
12 LL n;
13 bool mk[100010];
14 int p[100010],len;
15 inline LL LLabs(LL a){return a<0?-a:a;}
16 inline void getp()
17 {
18     for (int i=2;i<=100000;i++)
19     {
20         if (!mk[i])
21         {
22             p[++len]=i;
23             for (int j=2*i;j<=100000;j+=i)mk[j]=1;
24         }
25     }
26 }
27 inline void work()
28 {
29     int flag=(n<0),ans=0;n=LLabs(n);
30     for (int i=1;i<=len;i++)
31     {
32         if ((LL)p[i]*p[i]>n)break;
33         if (n%p[i]!=0)continue;
34         int now=0;while (n%p[i]==0)n/=p[i],now++;
35         if (!ans)ans=now;else ans=__gcd(ans,now);
36     }
37     if (n!=1)ans=1;
38     if (flag)while (ans%2==0)ans/=2;
39     printf("%d\n",ans);
40 }
41 int main()
42 {
43     getp();
44     while (~scanf("%lld",&n)&&n)work();
45 }

poj 1730

posted on 2017-08-04 15:22 zhber 阅读(151) 评论(0) 编辑收藏举报