1.当B的长度大于A时，直接将B前面的复制过去
2.当A的长度大于B时，B前面要补0，并且要判断奇数位偶数位
3.当A为10000,B为1时，要判断为0时应该打印0，最后打印10001，之前没判断直接A的那位数变负数再加10，所以会出问题
#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int main()
{char a[105],b[105],c[105];cin >> a >> b;int i = strlen(a)-1;int j = strlen(b)-1;//cout << i << j<<endl; int count=0;while(i>=0&&j>=0){if(count % 2 == 0){int x = (a[i]-'0' + b[j]-'0')%13;// cout << " x " <<x << endl;switch(x){case 10:c[count++] = 'J';break;case 11:c[count++] = 'Q';break;case 12:c[count++] = 'K';break;default:c[count++] = x+'0'; break;}i--;j--;// cout << "count "<<count<<endl;}else{int y;y = (b[j]-'0') - (a[i] - '0');if(y<0)y += 10;// cout << " " << y << endl; c[count++] = y + '0'; i--;j--; }}while(j>=0){c[count++] = b[j];j--; }while(i>=0){if(count % 2 == 0){c[count++] = a[i];}else{if(a[i]=='0')c[count++] = a[i];elsec[count++] = -(a[i]-'0')+10+'0'; }i--;}for(i=strlen(c)-1;i>=0;i--)cout << c[i];cout << endl;
}