密码工程-扩展欧几里得算法

密码工程-扩展欧几里得算法

0. 在openEuler(推荐)或Ubuntu或Windows(不推荐)中完成下面任务
1. 参考《密码工程》p112伪代码实现ExtendedGCD（int a, int b, int *k, int *u, int *v）算法（5
2. 在测试代码中计算74模167的逆。（5
3. 自己设计至少两个测试代码。5
4. 提交代码和运行结果截图

include <stdio.h>

//这里用了int类型，所支持的整数范围较小且本程序仅支持非负整数，可能缺乏实际用途，仅供演示。
struct EX_GCD { //s,t是裴蜀等式中的系数，gcd是a,b的最大公约数
int s;
int t;
int gcd;
};

struct EX_GCD extended_euclidean(int a, int b) {
struct EX_GCD ex_gcd;
if (b == 0) { //b等于0时直接结束求解
ex_gcd.s = 1;
ex_gcd.t = 0;
ex_gcd.gcd = 0;
return ex_gcd;
}
int old_r = a, r = b;
int old_s = 1, s = 0;
int old_t = 0, t = 1;
while (r != 0) { //按扩展欧几里得算法进行循环
int q = old_r / r;
int temp = old_r;
old_r = r;
r = temp - q * r;
temp = old_s;
old_s = s;
s = temp - q * s;
temp = old_t;
old_t = t;
t = temp - q * t;
}
ex_gcd.s = old_s;
ex_gcd.t = old_t;
ex_gcd.gcd = old_r;
return ex_gcd;
}

int main(void) {
int a, b;
printf("Please input two integers divided by a space.\n");
scanf("%d%d", &a, &b);
if (a < b) { //如果a小于b，则交换a和b以便后续求解
int temp = a;
a = b;
b = temp;
}
struct EX_GCD solution = extended_euclidean(a, b);
printf("%d%d+%d%d=%d\n", solution.s, a, solution.t, b, solution.gcd);
printf("所以%d模%d的逆=%d\n", a,b,solution.t);
return 0;
}