yt15-hdu-the least common multiple of the remainder theorem in China

Problem Description

I know that some of my classmates are looking at the Chinese remainder theorem recently, and this theorem itself is relatively simple:
Assuming that the m1,m2,..., Mk 22, the following equation group is the same:
X≡A1 (mod M1)
X≡A2 (mod m2)
...
X≡ak (MoD mk)
There is a unique solution within the 0<=<m1m2...mk.
Kee Mi=m/mi (1<=i<=k), because (mi,mi) = 1, so there are two integers pi,qi meet Mipi+miqi=1, if you remember Ei=mi/pi, then there will be:
Ei≡0 (mod MJ), j!=i
Ei≡1 (mod MJ), j=i
Obviously, E1a1+e2a2+...+ekak is a solution of the equations, and this solution adds and subtract an integer multiple of M to get the least nonnegative integer solution.
This is the Chinese remainder theorem and its solution process.
Now there is a problem with this:
A positive integer n divided by M1 (m1-a), divided by M2 remainder (m2-a), divided by M3 remainder (m3-a), in short, divided by MI remainder (mi-a), where (a<mi<100 i=1,2,... I) to find the minimum number of conditions to satisfy. 

Input

The input data contains multiple sets of test instances, the first line of each instance being two integers I (1<i<10) and a, where I represents the number of m, the meaning of a as described above, followed by a line of I integer M1,m1 ... Mi,i=0 and a=0 end input, not processed.

Output

For each test instance, output the minimum number of satisfied conditions in a row. The output for each instance takes up one row.

Sample Input

2 12 30 0

Sample Output

5

The code is as follows:

#include <iostream>using namespace Std;long long gcd (long long A, long  long B)         //Request Greatest common divisor {    long long t;
   if (A < b)     {        t=a;        a=b;        b=t;    }    if (b = = 0)        return A;    t= a%b;    while (t!= 0)    {        a=b;        b=t;        t=a%b;    }    return b;} int main () {    long long i,j,a,sum;    int str[10];    while (Cin>>i>>a)    {        if (i==0| | a==0) break            ;        Sum=1;        for (j=0; j<i; j + +)        {            cin>>str[j];            Sum= (Sum*str[j])/GCD (Sum,str[j]);          The least common multiple of two numbers = the product of two numbers/greatest common divisor           }        sum-=a;                                        Originally Yu (m-a) But at this time is the M to calculate, after the end should be subtracted from a                                          cout<<sum<<endl;    }    return 0;}

Operation Result:

yt15-hdu-the least common multiple of the remainder theorem in China