第八周OJ5打印数字图形

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问题及代码

/* 烟台大学计算机与控制工程学院 文件名称:打印数字图形 作    者:展一 完成时间:2016年10月20日 题目描述        从键盘输入一个整数n(1≤n≤9),打印出指定的数字图形。 输入        正整数n(1≤n≤9)。输出      指定数字图形。样例输入       5样例输出     1   121  12321 1234321123454321 1234321  12321   121    1*/  #include <stdio.h>#include <stdlib.h>int main(){    int n,i,j;    scanf("%d",&n);    for(i=1;i<=n;i++)    {        for(j=1;j<=n-i;j++)        printf(" ");        for(j=1;j<=i;j++)        printf("%d",j);        for(j=i-1;j>0;j--)        printf("%d",j);    printf("\n");    }    for(i=n-1;i>0;i--)    {        for(j=1;j<=n-i;j++)        printf(" ");        for(j=1;j<=i;j++)        printf("%d",j);        for(j=i-1;j>0;j--)        printf("%d",j);    printf("\n");    }    return 0;}


 

运行结果

知识点总结

循环结构的熟练运用。

以3为例,可以从最简单的结构开始,即1                    然后将空格输出          1              再根据循环将每一行的顺序数字输出                   1              

                                                               1 1                                                  1 1                                                                                       1 2

                                                               1 1 1                                            1 1 1                                                                                    1 2 3

然后根据                                               输出另一半图形          1             再将这一半程序copy下来,将for(i=1;i<=n;i++)改为for(i=n-1;i>0;i--)进行倒着的循环,输出    1

#include <stdio.h>                                                                1 2 1                                                                                                                                                        1 2 1
#include <stdlib.h>                                                            1 2 3 2 1                                                                                                                                                  1 2 3 2 1

int main()                                                                                                                                                                                                                                              1 2 1
{                                                                                                                                                                                                                                                               1
    int n,i,j;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n-i;j++)
        printf(" ");
        for(j=1;j<=i;j++)
        printf("%d",j);
        for(j=i-1;j>0;j--)
        printf("%d",j);
    printf("\n");
    }
    return 0;
}

 

学习心得

一开始自己做一点也不会,复习了两遍贺老师课上讲的从最初级开始单步思考逐渐复杂,开始弄懂一半,然而却不懂下一半的原理,之后看了贺老师的程序,仔细思考后懂了另一半的原理,并将程序完善。

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