ZOJ 3609 Modular Inverse 解线性模方程

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Modular Inverse

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Time Limit: 2 Seconds       Memory Limit: 65536 KB

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The modular modular multiplicative inverse of an integer a modulo m is an integer x such that a-1≡x (mod m). This is equivalent to ax≡1 (mod m).

Input

There are multiple test cases. The first line of input is an integer T ≈ 2000 indicating the number of test cases.

Each test case contains two integers 0 < a ≤ 1000 and 0 < m ≤ 1000.

Output

For each test case, output the smallest positive x. If such x doesn't exist, output "Not Exist".

Sample Input

3
3 11
4 12
5 13

Sample Output

4
Not Exist
8

模线性方程ax=b (mod n)，令d=exgcd(a,n),该方程有解的充要条件为 d | b ，即 b% d==0

方程ax=b(mod n)的最小解 :x=(x*(b/d))%n 

方程ax=b(mod n)的最整数小解: x=(x%(n/d)+n/d)%(n/d)

因为要求输出最小整数，所以如果答案为0的话，肯定是m=1的情况，此情况应输出1.

//0ms	168k
#include<stdio.h>
#include<string.h>
int exgcd(int A,int &x,int B,int &y)
{
    int x1,y1,x0,y0;
    x0=1;y0=0;
    x1=0;y1=1;
    int r=(A%B+B)%B;
    int q=(A-r)/B;
    x=0;y=1;
    while(r)
    {
        x=x0-q*x1;
        y=y0-q*y1;
        x0=x1;
        y0=y1;
        x1=x;y1=y;
        A=B;B=r;r=A%B;
        q=(A-r)/B;
    }
    return B;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int a,b=1,n,x,y;
        scanf("%d%d",&a,&n);
        int d=exgcd(a,x,n,y);
        if(b%d==0)
        {
            x=(x%(n/d)+n/d)%(n/d);
            if(!x)x++;
            printf("%d\n",x);
        }
        else printf("Not Exist\n");
    }
}