【整形数字转化为字符串，求有几位相同（汉明距离）】2021-11-20 20：15

缘由整形数字转化为字符串，求有几位相同（汉明距离）-编程语言-CSDN问答

void 汉明距离()
{int a = 0, aa = 0, j = 0;std::cin >> a >> aa;while (a && aa)(a % 10 == aa % 10 ? 0 : ++j), a /= 10, aa /= 10;if (a || aa)std::cout << "数据位数不同\n";else std::cout << j << std::endl;
}
void IntToStr(int n, char *str)
{//缘由https://ask.csdn.net/questions/7570982?spm=1005.2025.3001.5141int a = 0;while (n)str[a++] = n % 10 + '0', n /= 10;//char str2[10]{}, str1[10]{};//int n = 0;std::sort(str1, str1 + 10);std::sort(str2, str2 + 10);//scanf_s("%d", &n); /*std::cin >> n;*/ IntToStr(n, str1);//scanf_s("%d", &n); /*std::cin >> n;*/ IntToStr(n, str2);//printf_s("%d", differ(str1, str2)); /*std::cout << differ(str1, str2) << std::endl;*/
}
int differ(char *str1, char *str2)
{//求几个不同为何要转为字符串？不用转为字符串也可以实现！int a = 0, j = 0;while (str1[a] != '\0')if (str1[a] != str2[a])++j, ++a;else++a;return j;
}