How to solve the inverse problem of PHP multiplication
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Let S (x) denote the sum of factors of x. The title is: s (2004 ^ X) mod 29
Factor and S are product functions, that is, they satisfy property 1.
Property 1: if gcd (a) b) = 1, then S (a) = S (a) * S (b)
2004 ^ X = 4 ^ X * 3 ^ X * 167 ^ X
S (2004 ^ X) = S (2 ^ (2X)) * S (3 ^ X) * S (167 ^ X)
Property 2: if p is a prime, then S (p ^ X) = 1 ~ p + p ^ 2 +. + p ^ X = (p ^ (X ^ 1)-1) / (p ^ 1)
Therefore: s (2004 ^ X) = (2 ^ (2X+1)-1) * (3 ^ (XIX 1)-1) / 2 * (167 ^ (XIX 1)-1) / 166
167% 29 = 22
S (2004 ^ X) = (2 ^ (2X+1)-1) * (3 ^ (Xuan 1)-1) / 2 * (22 ^ (Xuan 1)-1) / 21
Nature 3: (aqb) / c% M = a% M * b% M * inv (c)
Where inv (c) is the smallest integer that satisfies (c*inv (c))% M é 1, where M = 29
Then inv (1) = 1 in v (2) = 15 in v (22) = 15
There are some good ones:
S (2004 ^ X) = (2 ^ (2X+1)-1) * (3 ^ (Xuan 1)-1) / 2 * (22 ^ (Xuan 1)-1) / 21
= (2 ^ (2X+1)-1) * (3 ^ (Xuan 1)-1) * 15 * (22 ^ (Xuan 1)-1) * 18
Fast power modulus is to find the value of a ^ n mod b in O (logn). The principle of the algorithm is ab mod c = (a mod c) (b mod c) mod c 390MS
# includeusing namespace std;const int pow [] [3] = {{2pr 5je 32}, {3pr 4je 81}, {22pr 2484}}; / 2 ^ 5 > 29,3 ^ 4 > 29,22 ^ 2 > 29, used to find (b ^ I)% 29int PowMod29 (int x index) / / Fast Modulus Power {int ans=1; while (index > = pow [x] [1]) / / when the index is greater than this value, it will exceed 29 {ans= (ans* quotient [x] [2])% 29 / / so model 29. And multiply by the previous value! Index-=pow [x] [1];} while (index--) ans= (ans* accounts [x] [0])% 29 bump / multiply the remaining ones not exceeding 29! Remember that on the mold 29 (because it is possible to exceed 29) return ans;} int main () {int Xpart2 part 2 department 3 part 3 part 167; while (cin > > Xerogram Xerox parts 0) {part2=PowMod29 (0meme 2 million Xerox 1); part3=PowMod29 (1 recorder Xero1); part167=PowMod29 (2 recital Xerox 1); cout
