HDU 1452 Happy 2004(约数和定理)

Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1904    Accepted Submission(s): 1394

Problem Description

Consider 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.

 

Input

The 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.

 

Output

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

 

Sample Input

1100000

 

Sample Output

610

 

Source

ACM暑期集训队练习赛（六） 

 

POINT：

了解一下约数和定理就行，用一下费马求逆元。

#include <iostream>#include <stdio.h>using namespace std;#define  LL long longconst int p = 29;int qkm(int base,int mi){    base%=p;    int ans=1;    while(mi)    {        if(mi&1) (ans*=base)%=p;        (base*=base)%=p;        mi>>=1;    }    return ans;}int main(){    int x;    while(~scanf("%d",&x)&&x)    {        int a=qkm(2,2*x+1)-1;        int b=((qkm(3,x+1)-1)+p)%p*qkm(2,p-2);        int c=((qkm(167,x+1)-1)+p)%p*qkm(166,p-2);        printf("%d\n",a*b*c%p);    }    }