Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 2021 Accepted Submission (s): 1474
Problem Descriptionconsider A positive integer x,and let S is the sum of all positive integer divisors of 2004^x. Your job is to determine s modulo (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,, 167, 334, 501, 668, 1002 and 2004. Therefore s = 4704 and s modulo are equal to 6.
Inputthe input consists of several test cases. Each test case is contains a line with the integer x (1 <= x <= 10000000).
A test Case of X = 0 indicates the end of input, and should not being processed.
Outputfor each test case, in a separate line, please output the result of S modulo 29.
Sample Input1100000
Sample output610 number theory, reasoning is very troublesome: the main idea of the topic is to ask 2004^n all the factors of the sum. According to the unique decomposition theorem 2004= (2^2) *3*167, then 2004^n= (2^2n) * (3^n) * (167^n), and the conclusion: the factor of a factor and is an integrable function. Set F (x) for the factor of X and, then F (AB) =f (a) *f (b); F (2004^n) =f (2^2n) *f (3^n) *f (167^n). Continue on conclusion: If a number is a prime, then F (a^n) =1+a+a^2+a^3+.......a^n = (a^ (n+1)-1)/(A-1); considering 167%29=22, then f (2004^n) = (2^ (2n+1)-1) * (3^ (n+1)-1)/2 * (22^ (n+1)-1)/21; then upper inverse: (a*b/c)%mod=a%mod* B%MOD*INV (c); where INV (c) represents the smallest integer (C*INV (c))%mod=1. Mod=29, then INV (1) =1;INV (2) =15;INV (21) = 18; the primitive = (2^ (2n+1)-1) * (3^ (n+1)-1)%MOD*INV (2) * (22^ (n+1)-1) *INV (21) 15*18% 29=9---------------The original (2^ (2n+1)-1) * (3^ (n+1)-1) * (22^ (n+1)-1) *9%29;
#include <iostream> #include <bits/stdc++.h>using namespace std;typedef long long ll;int ans=0;const int mod= 29;ll Quick_mod (ll k,ll N) { ll res=1; while (n>0) { if (n&1) { res= (res*k)%mod; } K=k*k%mod; n>>=1; } res--; if (res<0) res+=29; return res;} int main () { ll n; while (~SCANF ("%lld", &n) &&n) { ans=quick_mod (2,2*n+1) *quick_mod (3,n+1) *quick_mod (22,n+1) * 9%mod; printf ("%d\n", ans); } return 0;}
Happy 2004 hdu1452
