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=2200;
char s[]={'0','1','2','3','4','5','6','7','8','9','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(vis1,0,sizeof(vis1));
        memset(vis2,0,sizeof(vis2));
        int len1=a.length();  //反转第一个串
        for(int i=0;i<len1/2;i++){
            char temp=a[i];
            a[i]=a[len1-1-i];
            a[len1-1-i]=temp;
        }
        for(int i=0;i<len1;i++){
            int pos;
            for(int j=0;j<36;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=0;i<len2/2;i++){
            char tt=b[i];
            b[i]=b[len2-i-1];
            b[len2-1-i]=tt;
        }
        for(int i=0;i<len2;i++){
            int pos;
            for(int j=0;j<36;++j){
                if(b[i]==s[j]){
                    pos=j; break;
                }
            }
            vis2[i]=pos;
        }
        int mx=max(len1,len2);
        int cnt=0;
        for(int i=0;i<mx;i++){
            int temp=vis1[i]+vis2[i]+cnt;
            if(temp>=n){
                cnt=1;
                ans[i]+=temp%n;
            }else{
                cnt=0;
                ans[i]+=temp;
            }
        }
        if(cnt==1){
            ans[mx]=1; ++mx;
        }
        string ss;
        for(int i=0;i<mx;i++){
            ss+=s[ans[i]];
        }
        for(int i=0;i<mx/2;i++){
            char yy=ss[i];
            ss[i]=ss[mx-i-1];
            ss[mx-i-1]=yy;
        }
        cout<<ss<<endl;
    }
    return 0;
}