hdu 1452 Happy 2004（快速冪模數）__數論

求2004^X的因子和模29，用sum(2004^X)表示2004^X的因子和，sum(2004^X)=sum(4^X*3^X*167^X)，sum()是積性函數，

則有sum(2004^X)%29=sum(4^X)*sum(3^X)*sum(167^X)%29

                                =sum(2^(2*X))*sum(3^X)*sum(167^X)%29=sum(2^(2*X))*sum(3^X)*sum(22^X)%29

sum(p^(X))=1+p+p^2+……+p^n=(p^(n+1)-1)/(p-1)

則有：sum(2^(2*X))*sum(3^X)*sum(22^X)%29=((2^(2*X)-1)*(3^(X+1)-1)/2*(22^(X+1)-1)/21)%29

又有：15*2%29=1，21*18%29=1，則有15是2模29的乘法逆元，18是21模29的乘法逆元

所以可得：

sum(2004^X)%29=((2^(2*X)-1)*(3^(X+1)-1)*15*(22^(X+1)-1)*18)%29

#include<iostream>#include<cstdio>#include<string.h>#include<algorithm>#include<math.h>#include<queue>#include<stack>#include<map>#include<vector>#define mm(a,b) memset(a,b,sizeof(a))using namespace std;const int inf=0x7ffffff;const double PI=acos(-1.0);const double eps=1e-8;const double e=2.7182818284590452354;//計算a^bmodn   int modexp(int a,int b,int n)   {       int ret=1;         while(b)       {           if(b&1) ret=ret*a%n;          a=(a*a)%n;          b>>=1;       }       return ret;   }     int main(){int n;while(~scanf("%d",&n),n){int ans1=modexp(2,2*n+1,29);int ans2=modexp(3,n+1,29);int ans3=modexp(22,n+1,29);int ans=((ans1-1)*(ans2-1)*(ans3-1)*15*18)%29;printf("%d\n",ans);}    return 0;}