Harmony is indispensible in our daily life and no one can live without
it----may be Facer is the only exception. One day it is rumored that repeat
painting will create harmony and then hundreds of people started their endless
drawing. Their paintings were based on a small template and a simple method of
duplicating. Though Facer can easily imagine the style of the whole picture, but
he cannot find the essential harmony. Now you need to help Facer by showing the
picture on computer.
You will be given a template containing only one kind of
character and spaces, and the template shows how the endless picture is
created----use the characters as basic elements and put them in the right
position to form a bigger template, and then repeat and repeat doing that. Here
is an example.
# #
# <-template
# #
So the
Level 1 picture will be
# #
#
# #
Level 2 picture will be
# # #
#
# #
#
# # #
# #
#
#
#
# # #
#
# #
# #
# #
输入：
The input contains multiple test cases.
The first line of each case is an
integer N, representing the size of the template is N*N (N could only be 3, 4 or
5).
Next N lines describe the template.
The following line contains an
integer Q, which is the Scale Level of the picture.
Input is ended with a
case of N=0.
It is guaranteed that the size of one picture will not exceed
3000*3000.
For each test case, just print the Level Q picture by using the given template.
3 # # # # # 1 3 # # # # # 3 4 OO O O O O OO 2 0
# #
#
# #
# # # # # # # #
# # # #
# # # # # # # #
# # # #
# #
# # # #
# # # # # # # #
# # # #
# # # # # # # #
# # # #
# #
# # # #
# #
#
# #
# # # #
# #
# # # #
# # # # # # # #
# # # #
# # # # # # # #
# # # #
# #
# # # #
# # # # # # # #
# # # #
# # # # # # # #
OO OO
O OO O
O OO O
OO OO
OO OO
O O O O
O O O O
OO OO
OO OO
O O O O
O O O O
OO OO
OO OO
O OO O
O OO O
OO OO
递归：
#include<stdio.h> int n; char temp[5][5]; char p[3001][3001]; void printblank(int q,int k,int l) { int sum=1; for (int i=0;i<q;i++) sum=sum*n; for (int i=0;i<sum;i++) for (int j=0;j<sum;j++) p[k+i][l+j]=‘ ‘; } void print(int q,int k,int l) { if (q==1) { for (int i=0;i<n;i++) for (int j=0;j<n;j++) p[k+i][l+j]=temp[i][j]; }else { int sum=1; for (int i=0;i<q;i++) sum=sum*n; for (int i=0;i<n;i++) for (int j=0;j<n;j++) { int ki=i*sum/n; int kj=j*sum/n; if (temp[i][j]==‘ ‘) printblank(q-1,k+ki,l+kj); else print(q-1,k+ki,l+kj); } } } int main() { int q; while(scanf("%d",&n)!=EOF) { getchar(); if (n==0) continue; for (int i=0;i<n;i++) { for(int j=0;j<n;j++) { scanf("%c",&temp[i][j]); } getchar(); } scanf("%d",&q); print(q,0,0); int sum=1; for (int i=0;i<q;i++) sum=sum*n; for (int i=0;i<sum;i++) { for (int j=0;j<sum;j++) printf("%c",p[i][j]); printf("\n"); } } return 0; }
题目1161：Repeater,布布扣,bubuko.com
原文：http://www.cnblogs.com/Secontao/p/3597955.html