参见: 中国剩余定理, 大衍求一术
逆: 给定整数a,有(a,m)=1,称ax=1(mod m)的一个解叫做a模m的逆。
int res = (5544 * p + 14421 * e + 1288 * i - d + LCM) % LCM;
(数论不好真是硬伤...〒▽〒)
GCD 与 LCM:
int least_common_multiple(int a, int b)
{
int tmp, m, n;
m = a;
n = b;
while(b != 0)
{
tmp = a % b;
a = b;
b = tmp;
}
// a is greatest common divisor.
return m*n/a;
}