PTA 乙级 1048 数字加密 (20分) C++

 

 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

#include<iostream>
#include<string>
#include<algorithm>

using namespace std;

int main() {
    long a = 0, b = 0;    //必须用long,否则测试点0无法通过
    int an = 0, bn = 0;
    int odd = 0, even = 0;
    char arr[] = { '0','1','2','3','4','5','6','7','8','9','J','Q','K' };
    string s;
    cin >> a >> b;
    for (int i = 1; b % 10 != 0 || a % 10 != 0; ++i) {
        an = a % 10;
        bn = b % 10;
        a /= 10;
        b /= 10;
        if (i % 2 != 0) {
            odd = (an + bn) % 13;
            s += arr[odd];
        }
        else {
            even = bn - an;
            if (even < 0) even += 10;
            s += to_string(even);
        }
    }
    reverse(s.begin(), s.end());
    cout << s;
    return 0;
}

 

 

 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

#include<iostream>
#include<string>
#include<algorithm>

using namespace std;

int main() {
    string a, b;    
    /*奇位数，偶位数*/
    int odd = 0, even = 0;
    /*1~13位*/
    char arr[] = "0123456789JQK";
    string s;
    cin >> a >> b;
    reverse(a.begin(), a.end());        //进行反转，从正整数个位开始处理
    reverse(b.begin(), b.end());
    if (a.length() > b.length())        //比较长度，补0位
        b.append(a.length() - b.length(), '0');
    else
        a.append(b.length() - a.length(), '0');
    for (int i = 0; i <a.length(); ++i) {
        if (i % 2 == 0) {                //i为偶数即为原数字的奇数位
            odd = (a[i]-'0' + b[i] - '0') % 13;
            s = arr[odd] + s;
        }
        else {
            even = (b[i] - '0') - (a[i] - '0');
            if (even < 0) even += 10;
            s = to_string(even) + s;
        }
    }
    cout << s;
    return 0;
}