# China Remainder Theorem (also called Sun Tzu's theorem)

Source: Internet
Author: User

Today, I am bored occasionally. When I see something called the "China Surplus Theorem", I think it is fun to write a summary of the question first. Suppose that a number (1) is divided by 3 and 2, (2) divide by 5 and 4, (3) divide by 7 and 6, and obtain the minimum integer lcm (5, 7) that meets the conditions) 35 35*2 = 70 just divided by 3 remainder 1 lcm (3, 7) is 21 21 21*1 = 21 just divided by 5 remainder 1 lcm (5, 3) 15 15*1 = 15 just divided by more than 7 1 next (70*2 + 21*4 + 15*6) % lcm (70, 21, 15) = 104 final, 104 is the required result. The following is A mathematical formula: Assume that a number M is divided by the remainder of A, B, and C respectively, and the minimum integer (A, B, c, a, B, and c are all positive integers. Step 1: first solve LCM (B, C) = A' and multiply A' by an appropriate positive integer Ka (from 1 ), A '* Ka % A = 1 is terminated once it is established, so that a' = A' * Ka can be used to calculate B', C' is defined as above Step 2: result of solving LCM (A, B, C): The final result can be expressed as (a' * A + B '* B + C "* c) % LCM (, b, C); below, use C ++ to implement the complete code [cpp] # include <iostream> using namespace std; int GCD (int a, int B) {int tmp; if (a <B) return GCD (B, a); while (B) {tmp = B; B = a % B; a = tmp;} return ;} int LCM (int a, int B) {return a * B/GCD (a, B);} void GET2 (int & K2, int K) {int I = 1; while (1) {if (K2% K = 1) break; else K2 * = ++ I;} int main () {int A, A1, A2, B, b1, B2, C, C1, C2, a, B, c, I, D; cout <"Enter three divisor and remainder respectively :"; cin> A> a; cin> B> B; cin> C> c; // obtain the minimum public multiples of A', B ', C 'a2 = A1 = LCM (B, C); B2 = B1 = LCM (A, C); C2 = C1 = LCM (A, B ); D = LCM (A1, A); // The minimum public multiple of the three numbers. // you can call the following methods to obtain the minimum public multiples: A', B ', and C' GET2 (A2, ); GET2 (B2, B); GET2 (C2, C); cout <"Required number:" <(A2 * a + B2 * B + C2 * c) % D <endl; return 0 ;}

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