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HDU1452 Happy 2004 （因子和）

2024-07-26 19:35:55   220人阅读

Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)



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）
 
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首先2004 = 2*2*3*167
然后，利用因子和是积性函数的性质（蒟蒻准备专门写一篇持续更新的有关积性函数证明及学习的文章）：
σ(2004^x) =  σ(2^2x) * σ (3^x) * σ（167^x) Mod 29
∵167≡22（Mod29）
故σ(2004^x) =  σ(2^2x) * σ (3^x) * σ（22^x) Mod 29
                   =  [2^(2x+1)-1][3^(x+1)-1]/2*[22^(x+1)-1]/21
因为2的模29乘法逆元为15 ，22的模29乘法逆元为18
故σ(2004^x)  =  [2^(2x+1)-1][3^(x+1)-1]*15*[22^(x+1)-1]*18
即可用快速幂求解
 1 #include<set> 2 #include<queue> 3 #include<vector> 4 #include<cstdio> 5 #include<cstdlib> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std;10 const int Mod = 29;11 #define Rep(i,n) for(int i=1;i<=n;i++)12 #define For(i,l,r) for(int i=l;i<=r;i++)13 14 int ans,x;15 16 int quickpow(int m,int n){17     int ans=1;18     while(n){19         if(n&1) ans=(ans*m)%Mod;20         m=(m*m)%Mod;21         n>>=1; 22     }23     return ans%Mod;24 }25 26 int main(){27     while(scanf("%d",&x),x){28         int ans=0;29         ans=(quickpow(2,2*x+1)-1)%Mod;30         ans=ans%Mod*(quickpow(3,x+1)-1)*15%Mod;31         ans=ans%Mod*(quickpow(22,x+1)-1)*18%Mod;32         printf("%d\n",ans%Mod);33     }34     return 0;35 }
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HDU1452 Happy 2004 （因子和）


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