hdu 1452 Happy 2004 (快速冪+模數乘法逆元）

標籤：

Problem DescriptionConsider a positive integer X,and let S be the sum of all positive integer divisors of 2004^X. Your job is to determine S modulo 29 (the rest of the division of S by 29).
Take X = 1 for an example. The positive integer divisors of 2004^1 are 1, 2, 3, 4, 6, 12, 167, 334, 501, 668, 1002 and 2004. Therefore S = 4704 and S modulo 29 is equal to 6. 

 

InputThe input consists of several test cases. Each test case contains a line with the integer X (1 <= X <= 10000000). 
A test case of X = 0 indicates the end of input, and should not be processed. 

 

OutputFor each test case, in a separate line, please output the result of S modulo 29. 

 

Sample Input1100000 

 

Sample Output610

 

   題意：找2014^x%29的值。

   快速冪和模數乘法逆元。乘法逆元：x=(1/b)%m. 求x的值。x*b=k*m+1,即k去最小正整數時，x也為整數。即b能被k*m+1整除。

 

  

 1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 int f1(int x,int y) 5 { 6     int ans=1; 7     while (y) 8     { 9         if (y&1) ans=ans*x%29;10         x=x*x%29;11         y>>=1;12     }13     return ans;14 }15 int f2(int x)16 {17     int i=1;18     while (i)19     {20         if ((29*i+1)%x==0) break;21         i++;22     }23     return (29*i+1)/x;24 }25 int main()26 {27     int x,a,b,c;28     while (~scanf("%d",&x))29     {30         if (!x) break;31         a=(f1(2,2*x+1)-1)%29;32         c=(((f1(3,x+1)-1)%29)*f2(2))%29;33         b=(((f1(22,x+1)-1)%29)*f2(21))%29;34         printf("%d\n",(a*b*c)%29);35     }36     return 0;37 }

 

hdu 1452 Happy 2004 (快速冪+模數乘法逆元）