HDU 1452 Fzu 1053 Happy 2004 (inverse function + Factor and function + large exponent modulus simplification formula)

<span id="Label3"></p><p class="pst"><p class="pst">Description</p></p>Consider 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).<br><br>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.<p class="pst"><p class="pst">Input</p></p>The input consists of several test cases. Each test case is contains a line with the integer x (1 <= x <= 10000000).<br><br>A test Case of X = 0 indicates the end of input, and should not being processed.<p class="pst"><p class="pst">Output</p></p>For each test case, in a separate line, please output the result of S modulo 29.<p class="pst"><p class="pst">Sample Input</p></p>1100000<p class="pst"><p class="pst">Sample Output</p></p>610<pre><pre><br><br>The factor of a number and is an integrable function</pre></pre><p><p>About the integrable function, i.e. F (ab) =f (a) *f (b), There are many integrable functions in number theory</p></p><p><p>To prove it:</p></p><p><p>S (x) represents the factor of X and.</p></p><p><p>If x can be divided into a, b (which must be a prime number), then S (x) =s (a) *s (b).</p></p><p><p>Why must be divided into prime, because a prime factor after 1 and its own, for a, b, is 1,a,1,b, then x=a*b,x factor only 1,a,,b,x these four numbers,</p></p><p><p>(1+a) * (1+b) =1+a+b+a*b. see, This is called a number factor and is an integrable function.</p></p><p><p></p></p><p><p><strong>(a*b)/c%m= a%m * b%m * INV (c)<br></strong>Where INV (c) satisfies the smallest integer (C*INV (c))%m=1</p></p><p><p></p></p><p><p>a^ (b)%c----->a (b%k (c) +k (c))%c (b>=k (c)) where K (c) is the Euler value of C.</p></p><pre class="brush:cpp;gutter:true;"><pre class="brush:cpp;gutter:true;">#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath >using namespace Std;typedef Long LL; ll Qpow (int X,int y) { ll b = 1; For (; y;y>>=1) { if (y&1) b = (b*x)%29; x= (x*x)%29; } b--; Return b<0?b+29:b;} int main () { int x; While (~scanf ("%d", &x) &&x) { LL a = 2*x+1;if (a>28) a%=28,a+=28; LL B = x+1;if (b>28) b%=28,b+=28; printf ("%d\n", qpow (2,a) *qpow (3,b) *qpow (22,b) *9%29); } Return 0;}</pre></pre><p><p>　　</p></p><p><p>HDU 1452 Fzu 1053 Happy 2004 (inverse function + Factor and function + large exponent modulus simplification formula)</p></p></span>