Problem Description
RXD is a good mathematician.
One day he wants to calculate:
∑i=1nkμ2(i)×⌊nki−−−√⌋

output the answer module 109+7.
1≤n,k≤1018
μ(n)=1(n=1)
μ(n)=(−1)k(n=p1p2…pk)
μ(n)=0(otherwise)

p1,p2,p3…pk are different prime numbers
 

 

Input
There are several test cases, please keep reading until EOF.
There are exact 10000 cases.
For each test case, there are 2 numbers n,k.
 

 

Output
For each test case, output "Case #x: y", which means the test case number and the answer.
 

 

Sample Input
10 10
 

 

Sample Output
Case #1: 999999937

 打表大法好啊！打表之后发现就是求n^k%MOD

记得对n先做预处理取模，否则快速幂也救不了啊

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<cstring>
 5 using namespace std;
 6 const long long MOD=1e9+7;
 7 
 8 long long quickmod(long long a,long long b,long long m) 
 9 { 
10     long long ans = 1; 
11     while(b)//用一个循环从右到左遍历b的所有二进制位 
12     { 
13         if(b&1)//判断此时b[i]的二进制位是否为1 
14         { 
15             ans = (ans*a)%m;//乘到结果上，这里a是a^(2^i)%m 
16             b--;//把该为变0 
17         } 
18         b/=2; 
19         a = a*a%m; 
20     } 
21     return ans; 
22 } 
23 
24 int main()
25 {
26     long long n,k;
27     int t=1;
28     while(~scanf("%lld%lld",&n,&k))
29     {
30         printf("Case #%d: ",t++);
31         n%=MOD;
32         printf("%lld\n",quickmod(n,k,MOD));
33     }
34     return 0;
35 }

 

打表程序如下：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 using namespace std;
 5 
 6 #define MOD 1000000000+7
 7 
 8 bool panduan (long long num)
 9 {
10     long long i;
11     for(i=2;i<=sqrt((double)num)+1;i++)
12     {
13         if(num%(i*i)==0)
14             return true;
15     }
16     return false;
17 }
18 
19 int main()
20 {
21     int n,k;
22     long long num;
23     long long res=0;
24     for(int n=1;n<=10;n++)
25     for(int k=1;k<=10;k++)
26     {
27         res=0;
28         num=pow((double)n,(double)k);
29         for(int i=1;i<=num;i++)
30         {
31             if(!panduan(i))
32                 res+=(long long)(sqrt((double)(num/i)));
33             res%=MOD;
34         }
35         printf("%d %d %lld\n",n,k,res);
36     }
37     return 0;
38 }

 每一行三个数字分别表示n,k,res