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
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）
