密码工程-扩展欧几里得算法
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 | #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); return 0; } |
华为云实现:
计算74模167的逆
可知是79
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】博客园社区专享云产品让利特惠,阿里云新客6.5折上折
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步