zoj 3609 Modular Inverse(求逆元)

最新推荐文章于 2019-04-03 19:06:54 发布

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最新推荐文章于 2019-04-03 19:06:54 发布

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分类专栏：【编程之美】文章标签： gcd

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本文介绍了一种计算模逆元的方法，并提供了一个C++程序实现。该程序通过扩展欧几里得算法求解整数a关于模m的乘法逆元x，即满足ax ≡ 1 (mod m)的最小正整数x。若不存在这样的x，则输出'NotExist'。

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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

Sample Output

4
Not Exist
8

扩展gcd

#include <stdio.h>
#include <iostream>
#include <math.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
int exgcd(int a,int b,int &x,int &y)
{
    if (b==0)
    {
        x=1;
        y=0;
        return a;
    }
    int ans=exgcd(b, a%b, x, y);
    int temp=x;
    x=y;
    y=temp-a/b*y;
    return ans;
}
int cal(int a,int b,int c)
{
    int x,y;
    int gcd=exgcd(a, b, x, y);
    if(c%gcd!=0)
    {
        return -1;
    }
    x*=c/gcd;
    b/=gcd;
    if (b<0)
    {
        b=-b;
    }
    int ans=x%b;
    if (ans<=0)
    {
        ans+=b;
    }
    return ans;
}

int main()
{
    int a,b;
    int t;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d",&a,&b);
        int ans=cal(a, b,1);
        if(ans==-1)
        {
            printf("Not Exist\n");
        }
        else
        {
            printf("%d\n",ans);
        }
    }
    return 0;
    
}