Chinese remainder theorem (Sun Tzu's theorem)

"Chinese remainder theorem" is a famous arithmetic work in the period of 5-6 century in China's northern and Southern dynasties, a "matter unknown number" in the book of grandchildren, a solution to the problem: Today there is no know its number, 33 of the remaining two,

55 of the remaining three, 77 of the remaining two. Asking for geometry? Answer: 23.

According to the above we can get a set of formulas:

X≡2 (mod 3)

x≡3 (mod 5)

X≡2 (mod 7)

That

X% 3 = 2

X% 5 = 3

X% 7 = 2

Then we set X = T1 * A + t2 * b + t3 * C

Continue to set

a≡1 (mod 3) b≡0 (mod 3) c≡0 (mod 3)

A≡0 (mod 5) b≡1 (mod 5) c≡0 (mod 5)

A≡0 (mod 7) b≡0 (mod 7) c≡1 (mod 7)

A = LCM (5,7) * p (a%3=1) b = LCM (3,7) *p (b%5=1) c = LCM (3,5) *p (c%7=1)

= 35 * 2 = 70 = 21 * 1 = 21 = 15 * 1 = 15

Then we can get T1 = 2, t2 = 3, T3 = 2 based on the corresponding color.

So X = 2a + 3b + 2c = 2 * 70 + 3 * 21 + 2 * 15 = 233

233 is a solution to X, the general solution is x = 233 + gcd (3,5,7) *k = 233 + 105*k

These A, B, C (hereinafter referred to as NI) how to ask (Divisor 3, 5, 7 collectively referred to as MI)

X≡y1 (mod M1)

X≡y2 (mod m2)

X≡y3 (mod m3)

...

X≡an (mod mn)

Set x = N1 * y1 + N2 *y2 + N3 * y3 + ... + Nn * yn;

NI can be divisible by M1, M2, M3 、.... 、 mi-1, MI + 1 、... 、 mn, but not divisible by mi and ni%mi = 1

Set ni = LCM (m1, M2, M3, ..., MN)/mi * x = mi * y + 1, requires NI to have X

The above formula can be converted to LCM (M1, M2, M3, ..., MN)/mi * x + (-mi) * y = 1

After the transformation the formula is to expand Euclid, so that you can find the X

Template:

intCRT (intA[],intB[],intN) {//A[i] = divisor, b[i] denotes remainder    intM =1, N, ans =0, x, y;  for(inti =0; I < n; i++) M*= A[i];//because the divisor is 22 coprime, so least common multiple is the product     for(inti =0; I < n; i++) {N= M/A[i]; GCD (N, A[i], x, y);//using the extended Euclidean to find xAns + = N * x *B[i]; }    returnAns%M;}

Chinese remainder theorem (Sun Tzu's theorem)