zoj3609 Modular Inverse（扩展欧几里德算法）

Modular Inverse Time Limit: 2 Seconds Memory Limit: 65536 KB

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

题意：已知a，m求最小的x满足ax≡1（mod m）。

分析：扩展欧几里德的模板题，不多说。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

void exgcd(int a, int b, int &d, int &x, int &y)
{
    if (!b){d=a; x=1; y=0;}
    else
    {
        exgcd(b, a%b, d, y, x);
        y-=x*(a/b);
    }
}

int main()
{
    int a,m,T;
    cin>>T;
    while (T--)
    {
        cin>>a>>m;
        int d,x,y;
        exgcd(a, m, d, x, y);
        if (d==1)//如果余数为1
        {
            while (x<=0) x+=m;//结果可能为负数
            cout<<x<<endl;
        }
        else cout<<"Not Exist"<<endl;
    }
    return 0;
}