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NC24292. [USACO 2016 Dec B]The Cow-Signal

描述

Bessie and her cow friends are playing as their favorite cow superheroes. Of course, everyone knows that any self-respecting superhero needs a signal to call them to action. Bessie has drawn a special signal on a sheet of M×N paper (1≤M≤10,1≤N≤10), but this is too small, much too small! Bessie wants to amplify the signal so it is exactly K times bigger (1≤K≤10) in each direction.
The signal will consist only of the '.' and 'X' characters.

输入描述

The first line of input contains M, N, and K, separated by spaces.
The next M lines each contain a length-N string, collectively describing the picture of the signal.

输出描述

You should output KM lines, each with KN characters, giving a picture of the enlarged signal.

示例1

输入:

5 4 2
XXX.
X..X
XXX.
X..X
XXX.

输出:

XXXXXX..
XXXXXX..
XX....XX
XX....XX
XXXXXX..
XXXXXX..
XX....XX
XX....XX
XXXXXX..
XXXXXX..

原站题解

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Pascal(fpc 3.0.2) 解法, 执行用时: 2ms, 内存消耗: 256K, 提交时间: 2020-08-05 17:58:07

type
  str=string; 
var
  m,n,p,i,j,a,b:longint;
  s:str;
begin
    readln(m,n,p);
    for i:=1 to m do
    begin
        readln(s);
        for a:=1 to p do
        begin
            for j:=1 to length(s) do
            begin
                for b:=1 to p do
                  write(s[j]);
            end;
            writeln;
        end;
    end;
end.

C++11(clang++ 3.9) 解法, 执行用时: 3ms, 内存消耗: 504K, 提交时间: 2020-08-06 09:18:30

#include<stdio.h>
char s[15][15];
int main(){	
	int n,m,k;
	scanf("%d %d %d",&n,&m,&k);
	for(int i=0;i<n;i++)scanf("%s",s[i]);
	for(int i=0;i<n;i++){
	for(int sb=0;sb<k;sb++){
	for(int j=0;j<m;j++){
	for(int p=0;p<k;p++){
	printf("%c",s[i][j]);
	}}printf("\n");}}
} 

Python3(3.5.2) 解法, 执行用时: 17ms, 内存消耗: 3448K, 提交时间: 2020-08-06 10:14:09

a=input().split(' ')
for b in range(int(a[0])):
    c=input()
    e=""
    for d in range(int(a[1])):
        e=e+c[d]*int(a[2])
    for f in range(int(a[2])):
        print(e)

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