Happy 2004
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1346 Accepted Submission(s): 977
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.
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.
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的所有因子的和 模上29的值；
解题思路：
首先根据算术基本定理（1），
若 n 的标准素因子分解表达式为 n = p1^e1 * p2^e2 * …… * pk^ek
设f(n) 为n 的所有因子之和，则有:
f(n) = (p1^(e1+1)-1)/(p1-1) * (p2^(e2+1)-1)/(p2-1) * ... * (pk^(ek+1)-1)/(pk-1)
因为2004 = 2^2 * 3 * 167，所以2004^x也肯定是2^(2*x) * 3^x * 167^x;
然后根据算术基本定理：
167 Mod 29 == 22 所以直接用22就行
((2^(2*x+1)-1)/(2-1) * (3^(x+1)-1)/(3-1) * (22^(x+1)-1)/(22-1)) Mod 29
设a = (2^(2*x+1)-1)/1;
设b = (3^(x+1)-1)/2 ;
设c = (22^(x+1)-1)/21);
我们只要分别求出 a Mod 29, b Mod 29, c Mod 29的值就行啦，因为他们带着除数所以 我们进行求解逆元,分别是
1Mod29的逆元 2Mod29的逆元 21Mod29的逆元，这个可以根据扩展欧几里得算法得到，求出之后，我们就进行快速幂 分别求出2^(2x+1)Mod 29 , 3^(x+1)Mod 29, 22^(x+1)Mod 29 的值，然后减一 与 前面所求的逆元进行相乘就ok了
最后输出的就是 a*b*c Mod 29，下面是详细代码：
#include <iostream>
using namespace std;
typedef long long LL;
LL quick_mod(LL a, LL b, LL m)
{
LL ans = 1;
while(b)
{
if(b & 1)
ans = (ans*a)%m;
b>>=1;
a = (a*a)%m;
}
return ans;
}
void exgcd(LL a, LL b, LL &x, LL &y)
{
if(b == 0)
{
x = 1;
y = 0;
return;
}
LL x1, y1;
exgcd(b, a%b, x1, y1);
x = y1;
y = x1 - (a/b)*y1;
}
int main()
{
LL n, a, b, c, y, _a, _b, _c;
while(cin>>n,n)
{
exgcd(1,29,_a,y);
exgcd(2,29,_b,y);
exgcd(21,29,_c,y);
_a = (_a+29)%29;
_b = (_b+29)%29;
_c = (_c+29)%29;
///cout<<167%29<<endl;
///cout<<_a<<" "<<_b<<" "<<_c<<endl;
a=(quick_mod(2,2*n+1,29)-1)*_a;
b=(quick_mod(3,n+1,29)-1)*_b;
c=(quick_mod(22,n+1,29)-1)*_c;
cout<<(a*b*c%29)<<endl;
}
return 0;
}
