UVALive 7324 ASCII Addition （模拟）

ASCII Addition

题目链接：

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 ```
##题意：用题所示的格式输入 A + B 的式子，要以相同格式输出结果.
##题解：对每个数字存一下，把输入串分割成若干个数字(或'+')，然后暴力判读入的每个数即可. 注意string没有赋初值时候不能直接对某个位置赋值，直接用重定义过的'+'连接字符即可.
##代码： ``` cpp #include #include #include #include #include #include #include #include #include #include #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;

}