HDU 1452 Happy 2004 快乐2004

HDU 1452 Happy 2004 快乐2004 ：http://acm.hdu.edu.cn/showproblem.php?pid=1452

题面：

Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1351    Accepted Submission(s): 982

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

   
   
   
   

    
    
    
    1
10000
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    6
10
   
   
   
   

题目大意：

求2004^x所有因子的和并对29求模之后的结果。

题目分析：

先对2004进行素因子分解，可以得到2004=2^2*3*167，则2004^x=2^(2x)*3^x*167^x； //素因子分解

由算数基本定理：Φ(n)为n的所有因子之和，则有：

由题意，可计算得：

根据上述公式，用快速幂求解即可；

其中用到的公式和定理如下：

(a/b)mod m=(b^-1*a)mod m    // b^-1指的是b的逆元，要用扩展欧几里得求解，t徒手求解过程如下：

332*x+29*y=1

332=11*29+13

29=2*13+3

13=4*3+1

3=3*1

即gcd(332,29)=1

1=13-4*3

 =13-4*(29-2*13)

 =-4*29+9*13

 =-4*29+9*(332-11*29)

 =9*332-103*29

则x=9;

(a+b)mod m=(a mod m+b mod m)mod m

(a^b)mod m=(a mod m)^b mod m

(a*b)mod m=((a mod m)*(b mod m))mod m

附快速幂代码：

long long quick_mod(long long a,long long b,long long M)
{
    long long ans=1;
    while(b)
    {
        if(b&1)
        {
            ans=(ans*a)%M;
            b--;
        }
        b/=2;
        a=a*a%M;
    }
    return ans;
}

代码实现：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const double eps = 1e-4;
const int M=29;

long long quick_mod(long long a,long long b)
{
    long long ans=1;
    while(b)
    {
        if(b&1)
        {
            ans=(ans*a)%M;
            b--;
        }
        b/=2;
        a=a*a%M;
    }
    return ans;
}

int main()
{
    int x;
    while(scanf("%d",&x)!=EOF)
    {
        if(x==0)
        break;
        int sum;
        int a=(quick_mod(2,2*x+1)-1)%29;
        int b=(quick_mod(3,x+1)-1)%29;
        int c=(quick_mod(22,x+1)-1)%29;
        sum=(9*a*b*c)%29;
        printf("%d\n",sum);
    }
    return 0;
}