TAOCP-1.2.1_扩充的欧几里得算法

TAOCP-1.2.1_扩充的欧几里得算法

@(algorithm)[TAOCP]

1.算法E（扩充的欧几里得算法）

1.1问题描述:

给定两个正整数m和n，我们计算它们的最大公因子d和两个整数a和b，使得\(am + bn = d\)

1.2 实现步骤:

E1. [初始化] 置\(a' \xleftarrow{} b \xleftarrow{} 1\), \(a \xleftarrow{} b' \xleftarrow{} 0\), \(c \xleftarrow{} m\), \(d \xleftarrow{} n\)
E2. [除] 设q和r分别是c除以d的商和余数。（我们有 \(c = qd + r\), 且 \(0 \leq r \le d\)）
E3. [余数为0？] 如果r = 0，算法终止；在此情况下，我们如愿地有 \(am + b n = d\)
E4. [循环] 置\(c \xleftarrow{} d\), \(d \xleftarrow{} r\), \(t \xleftarrow{} a'\), \(a' \xleftarrow{} a\), \(a \xleftarrow{} t\) - \(qa\), \(t \xleftarrow{} b'\), \(b' \xleftarrow{} b\), $b \xleftarrow{} t $ - $ qb$，并返回E2

1.3 C实现

#include<stdio.h>
#include<stdint.h>

int32_t cycles;

int exgcd(int m,int n)
{
    uint32_t a, a1, b, b1, c, d;
    uint32_t q, r;
    uint32_t t;

    a1 = b = 1;
    a = b1 = 0;
    c = m;
    d = n;
    
    r = c % d;
    q = (c - r) / d; 
    
    while (r) {
        cycles++;
        c = d;
        d = r;
        t = a1;
        a1 = a;
        a = t -q * a;
        t = b1;
        b1 = b;
        b = t - q * b;
        
        r = c % d;
        q = (c - r) / d;
        if (r == 0)
            break;
    }
    return d;
}

int main(void)
{

    uint32_t m, n;
    uint32_t ret;
    
    printf("Please input 2 positive number(eg: 1769 551):\n");
    scanf("%d %d", &m, &n);

    ret = exgcd(m, n);

    printf("The gcd of %d and %d is: %d, cycles: %d\n", m, n, ret, cycles);

    return 0;
}

1.4 运行结果

# make
gcc -Wall -O3 -g e.c -o e
# ./e 
Please input 2 positive number(eg: 119 544):
1769 551
The gcd of 1769 and 551 is: 29, cycles: 3