B进制加法(洛谷1604)

分析：码农题，照这模拟就行，高精度的B进制，注意字符串反转的技巧。

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <string>

 #include <vector>

 #include <algorithm>

 #include <set>

 #include <map>

 #include <bitset>

 #include <cmath>

 #include <queue>

 #include <stack>

 using namespace std;

 const int maxn=;

 char s[]={'','','','','','','','','','','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};

 int n; //进制

 int vis1[maxn],vis2[maxn];

 int ans[maxn];

 int main()

 {

     while(cin>>n)

     {

         string a,b;

         cin>>a;

         cin>>b;

         memset(vis1,,sizeof(vis1));

         memset(vis2,,sizeof(vis2));

         int len1=a.length();  //反转第一个串

         for(int i=;i<len1/;i++){

             char temp=a[i];

             a[i]=a[len1--i];

             a[len1--i]=temp;

         }

         for(int i=;i<len1;i++){

             int pos;

             for(int j=;j<;j++){

                 if(a[i]==s[j]){

                     pos=j; break;

                 }

             }

             vis1[i]=pos;

         }

         int len2=b.length();   //反转第二个串

         for(int i=;i<len2/;i++){

             char tt=b[i];

             b[i]=b[len2-i-];

             b[len2--i]=tt;

         }

         for(int i=;i<len2;i++){

             int pos;

             for(int j=;j<;++j){

                 if(b[i]==s[j]){

                     pos=j; break;

                 }

             }

             vis2[i]=pos;

         }

         int mx=max(len1,len2);

         int cnt=;

         for(int i=;i<mx;i++){

             int temp=vis1[i]+vis2[i]+cnt;

             if(temp>=n){

                 cnt=;

                 ans[i]+=temp%n;

             }else{

                 cnt=;

                 ans[i]+=temp;

             }

         }

         if(cnt==){

             ans[mx]=; ++mx;

         }

         string ss;

         for(int i=;i<mx;i++){

             ss+=s[ans[i]];

         }

         for(int i=;i<mx/;i++){

             char yy=ss[i];

             ss[i]=ss[mx-i-];

             ss[mx-i-]=yy;

         }

         cout<<ss<<endl;

     }

     return ;

 }