Problem description
Xiao Ming designed a cross-shaped logo for an institution (not the Red Cross Society), as shown below:..$$$$$$$$$$$$$.. ..$...........$.. $$$.$$$$$$$$$.$$$ $...$.......$...$ $.$$$.$$$$$.$$$.$ $.$...$...$...$.$ $.$.$$$.$.$$$.$.$ $.$.$...$...$.$.$ $.$.$.$$$$$.$.$.$ $.$.$...$...$.$.$ $.$.$$$.$.$$$.$.$ $.$...$...$...$.$ $.$$$.$$$$$.$$$.$ $...$.......$...$ $$$.$$$$$$$$$.$$$ ..$...........$.. ..$$$$$$$$$$$$$..The other party also needs to output the logo in the form of characters in the computer dos window, and can control the number of layers arbitrarily.
Input
A positive integer n (n <30) indicates the number of layers required to print graphics.
Output
This flag corresponds to the number of surrounding layers.
Sample input
3Sample output
..$$$$$$$$$$$$$.. ..$...........$.. $$$.$$$$$$$$$.$$$ $...$.......$...$ $.$$$.$$$$$.$$$.$ $.$...$...$...$.$ $.$.$$$.$.$$$.$.$ $.$.$...$...$.$.$ $.$.$.$$$$$.$.$.$ $.$.$...$...$.$.$ $.$.$$$.$.$$$.$.$ $.$...$...$...$.$ $.$$$.$$$$$.$$$.$ $...$.......$...$ $$$.$$$$$$$$$.$$$ ..$...........$.. ..$$$$$$$$$$$$$..
Original title link: [Blue Bridge Cup] [The 4th Zhenti 2013] Print the cross
Simple graphic question, step-by-step analysis of the simulation process:
Step 1: Initialization (all '.')
int r = 5+n*4; //总共r行r列,5是中间的十字,加上左边2n,右边2n
for(int i=1;i<=r;i++){
for(int j=1;j<=r;j++){
t[i][j]='.';
}
}
result:
Step 2: Draw the cross in the middle (1/4 in the upper left corner)
for(int i=r/2+1-2;i<=r/2+1;i++){ //中间的十字 ,最中间的坐标为(r/2+1,r/2+1)
t[r/2+1][i]='$';
t[i][r/2+1]='$';
}
result:
Step 3: [Key] draw n layers around (1/4 in the upper left corner)
void quater(int r,int c,int w){
t[r][c]='$';
for(int i=1;i<=w;i++){ //向上走 w
t[--r][c]='$';
}
for(int i=1;i<=2;i++){ //向右走 2
t[r][++c]='$';
}
for(int i=1;i<=2;i++){ //向上走 2
t[--r][c]='$';
}
for(int i=1;i<=w;i++){ //向右走w
t[r][++c]='$';
}
}
{ ...
int row = r/2+1;
int col = r/2+1-4;
int walk=2;
while(n--){ //左上角四分之一,每个n为一层 ,起点为(row,col)
quater(row,col,walk);
col-=2;
walk+=2;
}
}
result:
Step 4: Symmetry
for(int j=r;j>r/2+1;j--) {
for(int i=1;i<=r/2+1;i++){
t[i][j] = t[i][r-j+1];
}
}
result:
Step 5: Symmetry up and down
for(int i=r;i>r/2+1;i--){
for(int j=1;j<=r;j++){
t[i][j] = t[r-i+1][j];
}
}
result:
So far, complete code is attached:
#include<cstdio>
#include<iostream>
using namespace std;
char t[150][150];
void quater(int r,int c,int w){
t[r][c]='$';
for(int i=1;i<=w;i++){ //向上走 w
t[--r][c]='$';
}
for(int i=1;i<=2;i++){ //向右走 2
t[r][++c]='$';
}
for(int i=1;i<=2;i++){ //向上走 2
t[--r][c]='$';
}
for(int i=1;i<=w;i++){ //向右走w
t[r][++c]='$';
}
}
int main(){
int n;
cin>>n;
int r = 5+n*4;
for(int i=1;i<=r;i++){
for(int j=1;j<=r;j++){
t[i][j]='.';
}
}
for(int i=r/2+1-2;i<=r/2+1;i++){ //中间的十字
t[r/2+1][i]='$';
t[i][r/2+1]='$';
}
int row = r/2+1;
int col = r/2+1-4;
int walk=2;
while(n--){ //左上角四分之一,每个n为一层 ,起点为(row,col)
quater(row,col,walk);
col-=2;
walk+=2;
}
//左右对称
for(int j=r;j>r/2+1;j--) {
for(int i=1;i<=r/2+1;i++){
t[i][j] = t[i][r-j+1];
}
}
//上下对称
for(int i=r;i>r/2+1;i--){
for(int j=1;j<=r;j++){
t[i][j] = t[r-i+1][j];
}
}
for(int i=1;i<=r;i++){
for(int j=1;j<=r;j++){
cout<<t[i][j];
}
cout<<endl;
}
return 0;
}