最大公约数-欧几米德,二进制方法

#include <iostream>using namespace std;//方法1：欧几米德公式int gcd1(int a, int b){if (b == 0){return a;}return gcd1(b , a % b);}//方法2：若一个整数同时能被a, b整除，那么一定也能够同时能被a-b，b整除//即gcd(a, b) = gcd(a - b, b) a > b//比如：gcd(42, 30) = gcd(12, 30) = gcd(30, 12) = gcd(18, 12) = gcd(6, 12) = gcd(12, 6) = gcd(6, 0) = 6//确点：比较次数增多了，效率低int gcd2(int a, int b){if (a < b){return gcd2(b, a);}if (b == 0){return a;}else{return gcd2(a - b, b);}}//方法3：二进制法//1）a,b同时为偶数那么gcd(a, b) = 2 * gcd(a / 2, b / 2)//2) a为偶数，b为奇数，那么gcd(a, b) = gcd(a / 2, b)//3) a为奇数，b为偶数，那么gcd(a, b) = gcd(a ， b/2)//4) a为奇数，b为奇数，那么gcd(a, b) = gcd(a - b, b)//效率：log(max(a, b))int gcd3(int a, int b){if (a < b){return gcd3(b , a);}if (b == 0){return a;}else{if ((a & 1) == 0){if ((b & 1 )== 0){return 2 * gcd3(a >>1, b >> 1);}else{return gcd3(a >>1, b);}}else{if ((b & 1) == 1){return gcd3(a-b, b);}else{return gcd3(a, b >> 1);}}}}int main(){int a = 42;int b = 30;printf("way1: %d \n", gcd1(a, b));printf("way2: %d \n", gcd2(a, b));printf("way3: %d \n", gcd3(a, b));return 0;}