最一开始想直接用整型，%10不断取余的处理方式，但是测试点4,5无法通过，原因在代码后面
1 #include<iostream> 2 #include<string> 3 #include<algorithm> 4 5 using namespace std; 6 7 int main() { 8 long a = 0, b = 0; //必须用long,否则测试点0无法通过 9 int an = 0, bn = 0; 10 int odd = 0, even = 0; 11 char arr[] = { '0','1','2','3','4','5','6','7','8','9','J','Q','K' }; 12 string s; 13 cin >> a >> b; 14 for (int i = 1; b % 10 != 0 || a % 10 != 0; ++i) { 15 an = a % 10; 16 bn = b % 10; 17 a /= 10; 18 b /= 10; 19 if (i % 2 != 0) { 20 odd = (an + bn) % 13; 21 s += arr[odd]; 22 } 23 else { 24 even = bn - an; 25 if (even < 0) even += 10; 26 s += to_string(even); 27 } 28 } 29 reverse(s.begin(), s.end()); 30 cout << s; 31 return 0; 32 }
以测试点5为例：
数字B的位数小于数字A的位数，而且数字A前面有0.
例如： A=000456 B=789
正确输出：000Q32
错误输出：:0:Q32
这一版的代码就会出现这样的毛病，测试点4同理，B前有0的话无法补全
所以只能用字符串处理了
1 #include<iostream> 2 #include<string> 3 #include<algorithm> 4 5 using namespace std; 6 7 int main() { 8 string a, b; 9 /*奇位数，偶位数*/ 10 int odd = 0, even = 0; 11 /*1~13位*/ 12 char arr[] = "0123456789JQK"; 13 string s; 14 cin >> a >> b; 15 reverse(a.begin(), a.end()); //进行反转，从正整数个位开始处理 16 reverse(b.begin(), b.end()); 17 if (a.length() > b.length()) //比较长度，补0位 18 b.append(a.length() - b.length(), '0'); 19 else 20 a.append(b.length() - a.length(), '0'); 21 for (int i = 0; i <a.length(); ++i) { 22 if (i % 2 == 0) { //i为偶数即为原数字的奇数位 23 odd = (a[i]-'0' + b[i] - '0') % 13; 24 s = arr[odd] + s; 25 } 26 else { 27 even = (b[i] - '0') - (a[i] - '0'); 28 if (even < 0) even += 10; 29 s = to_string(even) + s; 30 } 31 } 32 cout << s; 33 return 0; 34 }
