ZOJ 3609 Modular Inverse（扩展欧几里德）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4712

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

References

• http://en.wikipedia.org/wiki/Modular_inverse

代码如下:

#include <cstdio>
#include <cstring>
#include <cmath>
typedef long long LL;

LL exgcd(LL a,LL b,LL &x,LL &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    else
    {
        LL r = exgcd(b,a%b,x,y);
        LL t = x;
        x = y;
        y = t-a/b*y;
        return r;
    }
}

LL cal(LL a, LL b, LL c)
{
    LL x, y;
    LL tt = exgcd(a, b, x, y);
    if(c%tt)//无整数解
    {
        return -1;
    }
    x*=c/tt;
    b/=tt;
    if(b<0)
        b=-b;
    LL ans=x%b;
    if(ans<=0)
        ans+=b;
    return ans;
}

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