字符串的一道题

# include <bits/stdc++.h>
using namespace std;
const char *Const=" ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char s[101];
void Solve1() {
    int i=0,j=0,num1=0,num2=0;
    int t[101];
    for(++i;isdigit(s[i]);++i)
        (num2*=10)+=s[i]-'0';
    for(++i;s[i];++i)
        (num1*=10)+=s[i]-'0';
    for(;num1;num1=num1/26-!(num1%26)) {
        if(num1%26)
            t[++j]=num1%26;
        else
            t[++j]=26;    //特判
    }
    for(;j;--j)
        putchar(Const[t[j]]);
    printf("%d\n",num2);
}
void Solve2() {
    int i=0,num1=0,num2=0;
    for(;isupper(s[i]);++i)
        (num2*=26)+=s[i]-'A'+1;
    for(;s[i];++i)
        (num1*=10)+=s[i]-'0';
    printf("R%dC%d\n",num1,num2);
}
int main() {
    int n;
    bool flag;
    for(scanf("%d",&n);n;--n) {
        scanf("%s",s);
        flag=0;
        for(int i=0;s[i]&&!flag;++i)
            if(i&&isdigit(s[i-1])&&isupper(s[i]))
                flag=1;
        if(flag)
            Solve1();
        else
            Solve2();
    }
    return 0;
}

很简单易懂的题解。

#include <iostream>
#include <string>
#include <cmath>
#include <cctype>
using namespace std;
#define int long long
int n,a,b,fna;
string s,r,c;
bool isrxcy(string s){
	if(s.find("C")!=string::npos&&s.find("C")!=0&&isdigit(s[s.find("C")-1])){
		return true;
	}
	return false;
}
int szi(string s){
	int x=0;
	for(int i=0;i<s.length();i++){
		x*=10;
		x+=s[i]-'0';
	}
	return x;
}
string izs(int x){
	string s,s1;
	while(x!=0){
		char a[2]={'0'+x%10};
		s1=a;
		s.insert(0,s1);
		x/=10;
	}
	return s;
}
string rxcytorx(string a,int b){
	string s;
	while(b!=0){
		b--;
		char ca[2]={char(b%26+'A')};
		string bb=ca;
		s.insert(0,bb);
		b/=26;
	}
	return s+a;
}
string rxtorxcy(string a,int b){
	int x=0;
	for(int i=0;i<a.length();i++){
		x*=26;
		x+=a[i]-'A'+1;
	}
	return "R"+izs(b)+"C"+izs(x);
}
int findnumberaddress(){
	for(int i=0;i<s.length();i++){
		if(isdigit(s[i])){
			return i;
		}
	}
	return -1;
}
signed main(void){
	cin>>n;
	while(n--){
		cin>>s;
		if(isrxcy(s)){
			r=s.substr(1,s.find("C")-1);
			c=s.substr(s.find("C")+1,s.length()-1-s.find("C"));
			b=szi(c);
			r=rxcytorx(r,b);
		}else{
			fna=findnumberaddress();
			r=s.substr(0,fna);
			c=s.substr(fna,s.length()-r.length());
			a=szi(c); 
			r=rxtorxcy(r,a);
		}
		cout<<r<<endl;
	}
}