1B_十进制和二十六进制的相互转化

#include<iostream>
#include<string>
#include<queue>
#include<cmath>
#include<algorithm>
#include<cstring> 
#define ll long long
using namespace std;
void change_10to26(int s) {
    int len[100], cnt = 0; int temp;
    while (s) {
        temp = s % 26;
        if (temp == 0)temp = 26;//当temp为0的时候它就是Z
        len[cnt++] = temp;
        s = (s - temp) / 26;//为啥呢，比如如果输入，应该是A但不加这条就会出错.
    }
    for (int i = cnt - 1; i >= 0; i--)
        cout << (char)(len[i] + ‘A‘ - 1);
}
void change_26to10(string s) {
    int res = 0;
    for (int i = 0; i < s.size(); i++) {
        res *= 26; res += s[i] - 64;
    }
    cout << res;
}
int main() {
    int n;
    cin >> n;
    int i;
    while (n--) {
        char k[100];
        char op[100];
        scanf("%s", k);
        int a, b;
        if (sscanf(k, "R%dC%d", &a, &b) == 2) {//sscanf函数，如果没做到这题我是真的不知道，多学学多看看，我太菜了
            change_10to26(b);
            cout << a << endl;
        }
        else {
            string s = "";
            for (i = 0; k[i] >= ‘A‘&&k[i] <= ‘Z‘; i++)
                s += k[i];
            a = 0;
            for (int j = i; j < strlen(k); j++) {
                a *= 10;
                a += k[j] - ‘0‘;
            }
            cout << "R" << a << "C";
            change_26to10(s);
            cout << endl;
        }
    }
    return 0;
}