Description
This problem is a reverse case of the problem 2996. You are given the output of the problem H and your task is to find the corresponding input.
Input
according to output of problem 2996.
Output
according to input of problem 2996.
Sample Input
White: Ke1,Qd1,Ra1,Rh1,Bc1,Bf1,Nb1,a2,c2,d2,f2,g2,h2,a3,e4 Black: Ke8,Qd8,Ra8,Rh8,Bc8,Ng8,Nc6,a7,b7,c7,d7,e7,f7,h7,h6
Sample Output
+---+---+---+---+---+---+---+---+ |.r.|:::|.b.|:q:|.k.|:::|.n.|:r:| +---+---+---+---+---+---+---+---+ |:p:|.p.|:p:|.p.|:p:|.p.|:::|.p.| +---+---+---+---+---+---+---+---+ |...|:::|.n.|:::|...|:::|...|:p:| +---+---+---+---+---+---+---+---+ |:::|...|:::|...|:::|...|:::|...| +---+---+---+---+---+---+---+---+ |...|:::|...|:::|.P.|:::|...|:::| +---+---+---+---+---+---+---+---+ |:P:|...|:::|...|:::|...|:::|...| +---+---+---+---+---+---+---+---+ |.P.|:::|.P.|:P:|...|:P:|.P.|:P:| +---+---+---+---+---+---+---+---+ |:R:|.N.|:B:|.Q.|:K:|.B.|:::|.R.| +---+---+---+---+---+---+---+---+
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <string> using namespace std; char map[9][9]; void init() { memset(map, 0, sizeof(map)); string st; cin >> st; getline(cin, st); int i = 0; while (st[i] != '\0') { i++; char ch; if (st[i] > 'Z') ch = 'P'; else { ch = st[i]; i++; } int y = st[i++] - 'a' + 1; int x = st[i++] - '0'; x = 9 - x; map[x][y] = ch; } cin >> st; getline(cin, st); i = 0; while (st[i] != '\0') { i++; char ch; if (st[i] > 'Z') ch = 'p'; else { ch = st[i] + 'a' - 'A'; i++; } int y = st[i++] - 'a' + 1; int x = st[i++] - '0'; x = 9 - x; map[x][y] = ch; } } void output() { string line = "+---+---+---+---+---+---+---+---+"; cout << line << endl; for (int i = 1; i < 9; i++) { for (int j = 1; j < 9; j++) { char ch1 = map[i][j], ch; if ((i + j) % 2 == 1) ch = ':'; else ch = '.'; if (ch1 == 0) ch1 = ch; printf("|%c%c%c", ch, ch1, ch); } cout << "|" << endl; cout << line << endl; } } int main() { //freopen("D:\\t.txt", "r", stdin); init(); output(); return 0; }