wikioi 1276 图标的缩放



题目描述 Description

You have been asked to take a small icon that appears on the screen of a smart telephone and scale it up so it looks bigger on a regular computer screen.

你被要求把手机上的小图标变成电脑上的大的。

 

The icon will be encoded as characters (x and *) in a 3 × 3 grid as follows:

小图标是：

*x*

  xx

*  *
Write a program that accepts a positive integer scaling factor and outputs the scaled icon. A scaling factor of k means that each character is replaced by a k × k grid consisting only of that character.

写一个程序把它每一个字符都变成k*k的

输入描述 Input Description

The input will be a positive integer k such that k < 25.

输入会是一个正整数k，k<25。

输出描述 Output Description

The output will be 3k lines, which represent each individual line scaled by a factor of k and repeated k times. A line is scaled by a factor of k by replacing each character in the line with k copies of the character.

输出会是3k行，详见样例。

样例输入 Sample Input

3

样例输出 Sample Output

***xxx***

***xxx***

***xxx***

      xxxxxx

      xxxxxx

      xxxxxx
***     ***

***     ***

***     ***

数据范围及提示 Data Size & Hint

按照题目要求的缩进

#include<cstdio>#include<cstdlib>int main(){    int k;    scanf("%d",&k);    for(int j=0;j<k;j++)    {        for(int i=0;i<k;i++)            printf("*");        for(int i=0;i<k;i++)            printf("x");        for(int i=0;i<k;i++)            printf("*");        printf("\n");    }    for(int i=0;i<k;i++)    {        for(int m=0;m<k;m++)            printf(" ");        for(int j=0;j<k;j++)            printf("xx");        printf("\n");    }    for(int i=0;i<k;i++)    {        for(int j=0;j<k;j++)            printf("*");        printf(" ");        for(int p=0;p<k-2;p++)            printf(" ");        for(int m=0;m<k;m++)            printf("*");        printf("\n");    }    return 0;}

#include<cstdio>#include<cstdlib>char image[3][4]={"*x*"," xx","* *"};int main(){    int k;    scanf("%d",&k);    for(int i=0;i<3*k;i++)//3k*3k是宽*长，放大之前是3*3    {        for(int j=0;j<3*k;j++)        {            printf("%c",image[i/k][j/k]);        }        printf("\n");    }    return 0;}