标签：实现 letter 现在 back 行高 positive sts HERE bad
现在，如果你只是用手机的相机对着它们，智能手机应用可以即时翻译文本，甚至解决数学问题。您的工作是实现一个更简单的功能，回忆过去——添加两个作为ASCII艺术的整数。
ASCII艺术是一个字符矩阵，正好是7行高，每个字符都是点或小写字母X。
给出了A +B形式的表达式，其中A和B都是正整数。通过将所有的表达式字符（A和B的数字以及符号）作为7 5个矩阵，将这些矩阵转换成ASCII艺术，并将矩阵与单个字符的单个列串联在连续的各个矩阵之间。对应于数字和+符号的精确矩阵如下：
给定一个ASCII艺术来表达A+B的形式，找到加法的结果并用ASCII艺术形式写出。
输入由7行组成，包含用于A+B形式的表达式的ASCII技术，其中A和B都是由至多9个十进制数字组成的正整数，并且没有前导零。
输出包含ASCII艺术的7行，对应于加法的结果，没有前导零。
感谢@剑圣夜雨声烦 提供的翻译
Nowadays, there are smartphone applications that instantly translate text and even solve math problems if you just point your phone’s camera at them. Your job is to implement a much simpler functionality reminiscent of the past – add two integers written down as ASCII art.
An ASCII art is a matrix of characters, exactly 7 rows high, with each individual character either a dot or the lowercase letter x.
An expression of the form a + b is given, where both a and b are positive integers. The expression is converted into ASCII art by writing all the expression characters (the digits of a and b as well as the + sign) as 7 5 matrices, and concatenating the matrices together with a single column of dot characters between consecutive individual matrices. The exact matrices corresponding to the digits and the + sign are as follows:
Given an ASCII art for an expression of the form a + b, find the result of the addition and write it out in the ASCII art form.
输入格式：
Input consists of exactly 7 lines and contains the ASCII art for an expression of the form a + b, where both a and b are positive integers consisting of at most 9 decimal digits and written without leading zeros.
输出格式：
Output 7 lines containing ASCII art corresponding to the result of the addition, without leading zeros.
....x.xxxxx.xxxxx.x...x.xxxxx.xxxxx.xxxxx.......xxxxx.xxxxx.xxxxx ....x.....x.....x.x...x.x.....x.........x...x...x...x.x...x.x...x ....x.....x.....x.x...x.x.....x.........x...x...x...x.x...x.x...x ....x.xxxxx.xxxxx.xxxxx.xxxxx.xxxxx.....x.xxxxx.xxxxx.xxxxx.x...x ....x.x.........x.....x.....x.x...x.....x...x...x...x.....x.x...x ....x.x.........x.....x.....x.x...x.....x...x...x...x.....x.x...x ....x.xxxxx.xxxxx.....x.xxxxx.xxxxx.....x.......xxxxx.xxxxx.xxxxx
....x.xxxxx.xxxxx.xxxxx.x...x.xxxxx.xxxxx ....x.....x.....x.x.....x...x.x.........x ....x.....x.....x.x.....x...x.x.........x ....x.xxxxx.xxxxx.xxxxx.xxxxx.xxxxx.....x ....x.x.........x.....x.....x.....x.....x ....x.x.........x.....x.....x.....x.....x ....x.xxxxx.xxxxx.xxxxx.....x.xxxxx.....x
样例：1234567+890=1235457
Central Europe Regional Contest 2015 Problem A
大致是一道模拟题
（庆祝通过noip2018提高组初赛第五题）（五道黑题时代结束，之后开始正常练习）
1#include<bits/stdc++.h>
2#define int long long
3using namespace std;
4map<string,int> m;
5string st;
6string s[7];
7int n,row,line,loc,a,b,res,ans,w,ret;
8char out[10][5000];
9bool skip;
10void Init(){
11 st="xxxxxx...xx...xx...xx...xx...xxxxxx";
12 m.insert(pair<string,int>(st,0));
13 st="....x....x....x....x....x....x....x";
14 m.insert(pair<string,int>(st,1));
15 st="xxxxx....x....xxxxxxx....x....xxxxx";
16 m.insert(pair<string,int>(st,2));
17 st="xxxxx....x....xxxxxx....x....xxxxxx";
18 m.insert(pair<string,int>(st,3));
19 st="x...xx...xx...xxxxxx....x....x....x";
20 m.insert(pair<string,int>(st,4));
21 st="xxxxxx....x....xxxxx....x....xxxxxx";
22 m.insert(pair<string,int>(st,5));
23 st="xxxxxx....x....xxxxxx...xx...xxxxxx";
24 m.insert(pair<string,int>(st,6));
25 st="xxxxx....x....x....x....x....x....x";
26 m.insert(pair<string,int>(st,7));
27 st="xxxxxx...xx...xxxxxxx...xx...xxxxxx";
28 m.insert(pair<string,int>(st,8));
29 st="xxxxxx...xx...xxxxxx....x....xxxxxx";
30 m.insert(pair<string,int>(st,9));
31}
32void solve(){
33 for (int i=0;i<7;i++)
34 cin>>s[i];
35 n=s[0].length();
36 n=n/6; skip=true;n++;
37 for (int i=1;i<=n;i++){
38 loc=(i-1)*6;
39 st="";
40 for (int j=0;j<35;j++){
41 row=j/5; line=j%5;
42 st=st+s[row][loc+line];
43 }
44 if (skip){
45 if (!m.count(st)){
46 skip=false;
47 continue;
48 }
49 res=m[st];
50 a=a*10+res;
51 } else {
52 res=m[st];
53 b=b*10+res;
54 }
55 }
56 ans=a+b;
57}
58void writeln(string t,int num){
59 loc=(num-1)*5;
60 for (int i=0;i<35;i++){
61 row=i/5;line=i%5;
62 out[row][line+loc]=t[i];
63 }
64}
65void write(int num){
66 res=1;w=0;
67 while (res<=num){
68 w++;res*=10;
69 }
70 for (int i=1;i<=w;i++){
71 res/=10;
72 ret=num/res;ret%=10;
73 std::map<string,int>::iterator it;
74 for (it=m.begin();it!=m.end();it++)
75 if (it->second==ret)
76 writeln(it->first,i);
77 }
78 for (int i=0;i<7;i++){
79 for (int j=0;j<(w*5);j++){
80 cout << out[i][j];
81 if (j%5==4&&j!=w*5-1)
82 cout << ".";
83 }
84 cout << endl;
85 }
86}
87main(){
88 Init();
89 solve();
90 write(ans);
91 return 0;
92}
标签：实现 letter 现在 back 行高 positive sts HERE bad
原文地址：https://www.cnblogs.com/titititing/p/9822574.html
