題干：

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Happy 2004

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1863????Accepted Submission(s): 1361

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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.

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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.

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Output

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

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Sample Input

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1 10000 0

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Sample Output

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6 10

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Source

ACM暑期集訓(xùn)隊(duì)練習(xí)賽（六）

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題意：求2004^x的所有因子和對(duì)29取余的結(jié)果。

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題解：因?yàn)樗袛?shù)都可以被分解成a=p1^c1*p2^c2*...*pk^ck，并有約數(shù)和定理：sum=(p1^0+...+p1^c1)*(p2^0+...+p2^c1)*...*(pk^0+...+pk^ck)。每一項(xiàng)可以通過等比數(shù)列求和得到。2004=2^2*3*167，所以答案就是2^(2*n+1)*3^(n+1)*167^(n+1)/(2*166)。由于需要取余，所以不能直接除要求出2*166的逆元改為相乘。

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AC代碼：

#include<bits/stdc++.h> #define debug cout<<"aaa"<<endl #define mem(a,b) memset(a,b,sizeof(a)) #define LL long long #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define MIN_INT (-2147483647-1) #define MAX_INT 2147483647 #define MAX_LL 9223372036854775807i64 #define MIN_LL (-9223372036854775807i64-1) using namespace std;const int N = 100000 + 5; const int mod = 29;int e_gcd(int a,int b,int &x,int &y) {if(b==0) {x=1,y=0;return a;}int q=e_gcd(b,a%b,y,x);y-=a/b*x;return q;}int quick(int a,int b){int ans=1;while(b){if(b&1){ans=(ans*a)%mod;}b>>=1;a=(a*a)%mod;}return ans; }int main(){int n,x,y,a,b,c,ans;exgcd(166*2,29,x,y);while(~scanf("%d",&n)&&n){a=(quick(2,2*n+1)-1)%mod;b=(quick(3,n+1)-1)%mod;c=(quick(167,n+1)-1)%mod;ans=((a*b*c*x)%mod+mod)%mod;printf("%d\n",ans);}return 0; }

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