C语言实现求乘法逆元

如果ax≡1 (mod p),且gcd(a,p)=1（a与p互质），则称a关于模p的乘法逆元为x。

1.头文件

#include 
#include 
#include 

2.求需要存储的空间

int mod_inv_a(int a, int b, int *len)
{
    int d = 0;
    int i = 0;

    while(a)
    {
        d = b % a;
        b = a;
        a = d;
        i++;
    }

    *len = i;

    return 0;
}

3.求乘法逆元

int mod_inv_b(int a, int b, int len)
{
    int c[len];
    int d = 0;
    int e = 0;
    int f = 0;
    int g = b;
    int n = 0;
    int i = 0;

    memset(c, 0x00, sizeof(c));

    while(a)
    {
        d = b % a;
        c[i] = b / a;
        b = a;
        a = d;
        //printf("a=[%d], b=[%d], c=[%d]\n", a, b, c[i]);
        i++;
    }

    //printf("len=[%d]\n", len);
    d = 1;
    e = c[i-2];
    //printf("i-2=[%d]\n", i-2);
    n = i-1;
    i = 1;

    while(i < n)
    {
        //printf("i=[%d], n-1-i=[%d]\n", i, n-1-i);
        f = c[n-1-i] * e + d;
        f %= g;
        d = e;
        e = f;
        i++;
    }

    //e %= g;

    if(len % 2 == 0) e = g - e;

    return e;
}

4.接口封装

int mod_inv(int a, int b)
{
    int len = 0;

    mod_inv_a(a, b, &len);

    return mod_inv_b(a, b, len);
}

5.主函数部分

int main()
{
    int a = 19;
    int b = 97;
    int d = 0;

    d = mod_inv(a, b);

    printf("d=[%d]\n", d);
    printf("e=[%d]\n", d * a % b);

    return 0;
}