Congruence and modulo operation

1, basic operation:///Overflow condition, B is a positive integer

Addition: (a+b) mod n = ((a mod n) + (b mod n)) mod n

Subtraction: (-a) mod n = ((a mod n)-(b mod n)+n) mod n

Multiplication: AB mod n = (a mod n) (b mod n) mod n

2, large integer modulus:

    Char st[1000];    int m;    scanf ("%s%d", st,&m);    int Len =strlen (ST);    int ans=0;    for (int i=0;i<len;++i) {        ans= (int) ((Long Long) ans*10+st[i]-' 0 ')%m);    printf ("%d\n", ans);

3, Power modulo: a^n mod m

The use of Split-treatment method to solve

4, modular linear equations (with congruence): Ax = B (mod n) that is Ax-b = NY can be extended Euclidean algorithm solution

Special case: When B=1, the solution is a inverse of modulo n, when and only if the GCD of A and N (greatest common divisor is 1) there is a unique solution

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Congruence and modulo operation