ASCII Addition

題目鏈接：

http://acm.hust.edu.cn/vjudge/contest/127407#problem/A

Description

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 folows:

xxxxx ....x xxxxx xxxxx x...x 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...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.. xxxxx ....x xxxxx xxxxx ....x xxxxx xxxxx ....x xxxxx xxxxx .....

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.

Input

The input file contains several test cases, each of them as described below.
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.

Output

For each test case, output 7 lines containing ASCII art corresponding to the result of the addition,
without leading zeros.

Sample Input

....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

Sample Output

....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

題意：

用題所示的格式輸入 A + B 的式子，要以相同格式輸出結果.

題解：

對每個數字存一下，把輸入串分割成若干個數字(或'+')，然后暴力判讀入的每個數即可.
注意string沒有賦初值時候不能直接對某個位置賦值，直接用重定義過的'+'連接字符即可.

代碼：

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <queue> #include <map> #include <set> #include <vector> #include <list> #define LL long long #define eps 1e-8 #define maxn 1010 #define mod 100000007 #define inf 0x3f3f3f3f #define mid(a,b) ((a+b)>>1) #define IN freopen("in.txt","r",stdin); using namespace std;string num[11][7] = {{"xxxxx","x...x","x...x","x...x","x...x","x...x","xxxxx"},{"....x","....x","....x","....x","....x","....x","....x",},{"xxxxx","....x","....x","xxxxx","x....","x....","xxxxx"},{"xxxxx","....x","....x","xxxxx","....x","....x","xxxxx"},{"x...x","x...x","x...x","xxxxx","....x","....x","....x"},{"xxxxx","x....","x....","xxxxx","....x","....x","xxxxx"},{"xxxxx","x....","x....","xxxxx","x...x","x...x","xxxxx"},{"xxxxx","....x","....x","....x","....x","....x","....x",},{"xxxxx","x...x","x...x","xxxxx","x...x","x...x","xxxxx"},{"xxxxx","x...x","x...x","xxxxx","....x","....x","xxxxx"},{".....","..x..","..x..","xxxxx","..x..","..x..","....."} };int match(string cur[]) {for(int i=0; i<10; i++) {int flag = 1;for(int j=0; j<7; j++) {for(int k=0; k<5; k++)if(num[i][j][k] != cur[j][k]) {flag = 0; break;}}if(flag) return i;}return -1; }string data[10]; string print[10];int main(int argc, char const *argv[]) {//IN;while(cin >> data[0]){for(int i=1; i<7; i++)cin >> data[i];int len = data[0].size();int n = (len + 1) / 6;LL a=0,b=0; int flag = 0;int start = 0;string cur[10];for(int i=1; i<=n; i++) {for(int j=0; j<7; j++) {cur[j].clear();for(int k=start; k<start+5; k++) {cur[j] += data[j][k];}}int dig = match(cur);if(dig == -1) {flag = 1;start += 6;continue;}if(!flag) {a = a*10LL + dig;} else {b = b*10LL + dig;}start += 6;}LL Ans = a + b;vector<int> ans; ans.clear();while(Ans) {int di = Ans % 10LL;Ans /= 10LL;ans.push_back(di);}int sz = ans.size();for(int i=0; i<10; i++) print[i].clear();for(int i=sz-1; i>=0; i--) {for(int j=0; j<7; j++) {for(int k=0; k<5; k++) {print[j] += num[ans[i]][j][k];}if(i) print[j] += '.';}}for(int i=0; i<7; i++) {cout << print[i] << endl;}}return 0; }

轉載于:https://www.cnblogs.com/Sunshine-tcf/p/5757847.html

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