寒假刷题之7——波纹

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mb61b856e04bb98 2021-12-16 17:03:44 ©著作权

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Triangle Wave

In this problem you are to generate a triangular wave form according to a specified pair of Amplitude and Frequency.

Input and Output

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

Each input set will contain two integers, each on a separate line. The first integer is the Amplitude; the second integer is the Frequency.

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

For the output of your program, you will be printing wave forms each separated by a blank line. The total number of wave forms equals the Frequency, and the horizontal ``height'' of each wave equals the Amplitude. The Amplitude will never be greater than nine.

The waveform itself should be filled with integers on each line which indicate the ``height'' of that line.

NOTE: There is a blank line after each separate waveform, excluding the last one.

Sample Input

1
3
2

Sample Output

波纹。。。让我想起jojo里面的波纹（好吧我漫画看多了。。。）

其实这就是画菱形的变形题嘛，而且比画菱形容易。。。

编出来调试正常却ac不了：

#include <stdio.h>
#include <ctype.h>
int main()
{
  int i, k, j, l, n, f, s;

  scanf("%d", &n);

  for (i = 0; i < n; i ++){
    scanf("%d%d", &f, &s);
    for (l = 0; l < s; l ++){
      for (k = 1; k <= f; k ++){
        for (j = 0; j < k; j ++)
          printf("%d", k);
        printf("\n");
      }
      for (k = f - 1; k > 0; k --){
        for (j = k; j > 0; j --)
          printf("%d", k);
        printf("\n");
      }
      if (l != s - 1)
        putchar('\n');
    }
  }

  return 0;
}

原来是空行问题，调整一下就ac了。。。它特别要求最后一个没有空行，我以为每一组的最后一个都没有，原来是算全部啊。。。

那个oj好卡反应很慢很蛋疼。。。

AC代码：

#include <stdio.h>
#include <ctype.h>
int main()
{
  int i, k, j, l, n, f, s;

  scanf("%d", &n);

  for (i = 0; i < n; i ++){
    scanf("%d%d", &f, &s);
    for (l = 0; l < s; l ++){
      for (k = 1; k <= f; k ++){
        for (j = 0; j < k; j ++)
          printf("%d", k);
        printf("\n");
      }
      for (k = f - 1; k > 0; k --){
        for (j = k; j > 0; j --)
          printf("%d", k);
        printf("\n");
      }
      if (l == s - 1 && i == n - 1)
        return 0;
      putchar('\n');
    }
  }

  return 0;
}