2018/7/31-zznuoj-问题 A: A + B 普拉斯【二维字符串+暴力模拟+考虑瑕疵的题意-0的特例】

问题 A: A + B 普拉斯

在计算机中，数字是通过像01像素矩阵来显示的，最终的显示效果如下： 
 
现在我们用01来构成这些数字

当宝儿姐输入A + B 时(log10(A)<50，log10(B)<50,且A,B均为正整数)，你来计算A+B的和C，并按格式在屏幕上打印C。

输入

每组输入包括两个非负整数 A,B（log10(A)<50,log10(B)<50）,已EOF结束输入 

输出

按格式在屏幕中打印C，数字之间相隔三列0。 

样例输入

3 8

样例输出

0010000000100
0110000001100
0010000000100
0010000000100
0010000000100
0010000000100
0111000001110

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解析思路：

1、首先按照大数相加，把结果求出来！方法：开两个数组暴力模拟！

2、坑点：最后需要把大数的结果按照正常的顺序输出，并且是转化成01字符串！

3、01串事先存在一个二维字符串中，第一维表示0--9以及空格（三列零），第二维表示每个代表的数字（0--9以及空格）的横坐标，得到具体的横坐标后就一层一层地输出即可！把二维字符串想象成蛋糕，输出时就像横着切蛋糕一样，一层一层地去切！

4、题意，有点瑕疵！还有零的情况吧！比如零加零的情况！

---

题解：

 #include<iostream>
 #include<stdio.h>
 #include<string.h>
 #include<string>
 #include<vector>
 #include<algorithm>
 #define ll long long
 using namespace std;
 #define N 100

 char s1[],s2[];
 int sum[];
 vector<int>order;

 char num[][][]={
      {"","","","","",
 "",""},
     {"","","","","",
 "",""},
 {"","","","","",
                     "",""},
 {"","","","","",
 "",""},
 {"","","","","",
 "",""},
  {"","","","","",
 "",""},
  {"","","","","",
 "",""},
 {"","","","","",
 "",""},
  {"","","","","",
 "",""},
  {"","","","","",
 "",""},
 {"","","","","",
 "",""}};

 void paint( )
 {
     for(int i=;i<=;i++){
         for(int j=;j<(int)order.size();j++){
             int x=order[j];
             printf("%s",num[x][i]);
         }
         cout<<endl;
     }
 }

 int add(char s1[],char s2[]){
     int len1=strlen(s1);
     int len2=strlen(s2);
     int i1=len1-,i2=len2-,i3=;
     while(i1>=&&i2>=){
         sum[i3]+=s1[i1]-''+s2[i2]-'';  //记得这里是加等于!
         if(sum[i3]>=){
             sum[i3+]+=sum[i3]/;
             sum[i3]%=;
         }
         i3++,i1--,i2--;
     }
     while(i1>=){
          sum[i3]+=s1[i1]-'';
         if(sum[i3]>=){
             sum[i3+]+=sum[i3]/;
             sum[i3]%=;
         }
          i3++,i1--;
     }
     while(i2>=){
          sum[i3]+=s2[i2]-'';
         if(sum[i3]>=){
             sum[i3+]+=sum[i3]/;
             sum[i3]%=;
         }
          i3++,i2--;
     }
     while(sum[i3]>=){  //最终的进位情况!!
             sum[i3+]+=sum[i3]/;
             sum[i3]%=;
             i3++;
     }

     if(sum[i3]>)
         return i3+;
     else
         return i3;
 }

 int main()
 {
     while(scanf("%s%s",s1,s2)!=EOF){
         memset(sum,,sizeof(sum));
         int len3=add(s1,s2);   //计算大数相加

        /* for(int i=0;i<len3;i++){
             printf("%d",sum[i]);
         }
         cout<<"*****"<<endl;
 */
         order.clear();    //生成次序
         for(int i=len3-;i>=;i--){
             order.push_back(sum[i]);
             if(i!=)
                 order.push_back();
         }
         paint();
     }

     return ;
 }

（用到了vector来存贮结果）