HDU 1576 A/B (Extension of Euclidean Algorithm)

Topic link:

http://acm.hdu.edu.cn/showproblem.php?pid=1576

topic:

Problem Description

(A/B)%9973 is required, but since A is very large, we only give n(n=A%9973) (our given A must be divisible by B, and gcd(B,9973) = 1).

 

 

Input

The first row of data is a T, indicating that there are T groups of data.
Each set of data has two numbers n (0 <= n < 9973) and B (1 <= B <= 10^9).

 

 

Output

Corresponding to each group of data output (A/B)% 9973.

 

 

Sample Input

2

1000 53

87 123456789

 

Sample Output

7922

6060

1  /* 
2  Problem
 3  gives n (0 <= n < 9973) and B (1 <= B <= 10^9), calculate (A/B)%9973
 4  where n=A%9973, A must be able to Divisible by B, and gcd(B, 9973) = 1 
 5  
6 Problem-  solving ideas
 7  Set X=(A/B)%9973, then A/B=K*9973+X (K is a positive integer)
 8  is A= K*9973*B+X*B
 9  
10  and n=A%9973, then A=p*9973+n
 11  
12  Combining the two formulas can get p*9973+n=K*9973*B+X*B
 13  shift The item can be obtained as p*9973-K*9973*B=X*Bn
 14  , that is, 9973*(pK*B)=X*Bn
 15  Obviously, the remainder of 9973 on the left is equal to 0, then 0=(X*Bn)%9973 
 16  At this time, you can directly enumerate the value of X. In addition, it is noted that X*B may exceed the range of int, and the congruence operation needs to be used. Of course, it can also be directly converted to long long type. 
17  (ab)%n=((a%n)-(b%n)+n)%n
 18  (a*b)%n=(a%n)*(b%n)%n
19 故(x*B-n)%9973=( ((X%9973)*(B%9973)%9973) -(n%9973)+9973 )%9973
20 */
21 
22 #include<cstdio>
23 
24 int main()
25 {
26     int t,n,X;
27     long long B;
28     scanf("%d",&t);
29     while(t--){
30         scanf("%d%lld",&n,&B);
31         for(X=0;;X++){
32             if((((X%9973)*(B%9973)%9973) - (n%9973)+9973)%9973 == 0){
33                 printf("%d\n",X);
34                 break;
35             }
36         }
37     }
38     return 0;
39 }
40