Poj 2891 modulus linear equations (mi mj non-reciprocal)

Question:

X = AI (mod mi) 1 <= I <= K

Consider K = 2 first:

X = A1 (mod M1)

X = a2 (mod m2)

D | (a1-a2), where D = (M1, M2)

The proof is as follows:

Necessity: Let X be the solution of the above homogeneous equations, so there is an integer Q1, Q2 makes X = A1 + m1 * Q1, x = a2 + m2 * Q2, remove X and get a1-a2 = m2q2-m1q1. Because d = (M1, M2), d | (a1-a2 ).

Adequacy: If D = (M1, M2) | (a1-a2) is true, then m1 * x + m2 * Y = a1-a2 has a solution.

Set to x0 and Y0. So m2 * Y0 = a1-a2 (mod M1)

Note X1 = a2 + m2 * y0, you can know X1 = a2 (mod m2), and X1 = a2 + m2 * Y0 = a2 + (a1-a2) = A1 (mod M1)

So X1 = a2 (mod m2) = A1 (mod M1)

So X = x1 (mod [M1, M2])

In addition, if both X1 and X2 are solutions of the same equations above, X1 = x2 (mod M1), X1 = x2 (mod m2 ), from the same remainder, X1 = x2 (mod [M1, M2]) is obtained. That is, the interpretation of the modulus [M1, M2] and the same remainder equations is unique.

Pass.

····

K> 2. .

# Include <cstdio> # define lint _ int64lint egcd (lint A, lint B, lint & X, lint & Y) {lint TMP, RET; If (B = 0) {x = 1, y = 0; return a;} ret = egcd (B, A % B, x, y); TMP = x, x = Y, y = TMP-A/B * Y; return ret;} lint gcd (lint A, lint B) {If (B = 0) return a; return gcd (B, A % B);} // obtain the modulus linear equations Ax = B (mod m) lint modular_linear (lint A, lint B, lint m) {lint D, X, Y; D = egcd (a, m, X, Y); If (B % d! = 0) Return-1; // if no solution is available, return-1. Return (x * (B/d) % m + M) % m;} int main () {lint A1, A2, M1, M2, K, Y; while (scanf ("% i64d", & K )! = EOF) {bool flag = false; scanf ("% i64d % i64d", & M1, & A1); If (k = 1) y = A1; for (INT I = 1; I <K; I ++) {scanf ("% i64d % i64d", & m2, & A2); If (FLAG) continue; lint y = modular_linear (M2, A1-A2, M1); If (y =-1) {flag = true; continue;} a1 = a2 + m2 * Y; m1 = m1 * m2/gcd (M1, M2); a1 = (a1% M1 + M1) % M1;} If (FLAG) printf ("-1 \ n "); else printf ("% i64d \ n", A1);} return 0 ;}