/* Print the Following Pattern (Truth Table) ---> Unbeatable Challenge input= 1 test case Mandatory :: User Input All test cases : 1. 0+X=X 2. 1+X=1 3. 0.X=0 4. 1.X=X 5. X+X=X 6. X.X=X Contraint : Any alphabet can be taken. Alphabet must be in capital letter in a test case. All test cases can be given in input in reverse order. Like : 0+X=X or X+0=X Try to code in minimum lines Link your answers in comments No one has completed it yet Input = X.Y or X+Y or 1.X or 1.Y */ #include #include using namespace std; using VA = valarray; using IDX = valarray; // box dimensions constexpr size_t W{42}; constexpr size_t H{11}; // offset of chars to place in the table IDX const idx1{2*W+6, 2*W+18, 2*W+32}; IDX const idx2{6*W+6, 6*W+18, 6*W+32}; IDX const idx3{8*W+6, 8*W+18, 8*W+32}; void drawbox(VA& v) { v[slice(W-1,H-1,W)]='\n'; v[W*H-1]='\0'; v[gslice(0, {2, W / 2}, {4 * W, 2})] = '*'; v[gslice(0, {2, H},{W-2,W})]='*'; v[gslice(12, {2, H},{12,W})]='*'; v[slice((H -1) * W, W / 2, 2)] = '*'; } int main() { VA v(' ', W * H); drawbox(v); VA R(3); char l='X', r = '1', op = '+'; cin >> l >> op >> r; v[idx1] = VA{l, r, 'R'}; auto Result = [](auto&& R, char op) { if(op == '+') R[2] = (char) (R[0] | R[1]); else R[2] = (char) (R[0] & R[1]); }; auto Row = [&](auto && v, auto&& R, char l, char op, char r, IDX id) { R[0] = l, R[1] = r; Result(R, op); v[id] = R; }; cout << "Truth Table for "<< l<

[Challenge] Kiibo_Pattern20.cpp