在C中打印Pythagorean Triple的代表 [英] Print a representation of the Pythagorean Triple in C

问题描述

我试图创建一个程序，用于打印在C中找到的pythagorean三元组的映射数据。到目前为止，我已经编写了程序以便能够找到三元组。

  #include< stdio.h> 
 #include< math.h> 
 
 int main（int argc，char * argv []）{
 int a，b，c; 
 int a2，b2，c2; 
 int limit = 60; （a = 1; a< = limit; a ++）{
 a2 = a * a; 
 
; （b = 1; b <=极限; b ++）{
 b2 = b * b; （c = 0; c <=（（int）sqrt（a2 + b2））; c ++）{
 c2 = c * c; 
 if（a  printf（triple：％d％d％d \ n，a，b， C）; 
} 
} 
} 
} 
} 
  

预期输出格式为：

  123456789012345678901234567890123456789012345678901234567890 
 1 \ 
 2 \ 
 3 \ 
 4 * \ 
 5 \ 
 6 \ 
 7 \ 
 8 * \ 
 9 \ 
 0 \ 
 1 \ 
 2 * * \ 
  

我正在尝试写一个这样的循环，但无法想象如何以这种方式进行打印。任何建议？

更新：我设法打印x和y轴（x = a和y = b）。

  for（int x = 0; x  printf（％d，x）; 
} 
 
 printf（\\\
）; 
 
 for（int y = 0; y  
 printf（％d \ n，y） ; 
 
  

更新：修改代码，输出是打印，但打印空间有问题。尝试手动为\和*添加空格，但这只是扭曲了整个图像。

  #include< stdio.h> 
 #include< math.h> 
 #include< stdbool.h> 
 
 bool is_perfect_square（int num）; 
 
 int main（int argc，char * argv []）{
 
 int a，b，c; 
 int a2，b2，c2; 
 int limit = 60; 
 bool flag = false; （a = 1; a< = limit; a ++）{
 a2 = a * a; 
 
; （b = 1; b <=极限; b ++）{
 b2 = b * b; （c = 0; c <=（（int）sqrt（a2 + b2））; c ++）{
 c2 = c * c; 
 if（a  // printf（triple：％d％d％d \\\
，a， b，c）; （int x = 0; x  for（int）
 
 
 
 
 $ b y = 0; y  if（x == 0）{
 printf（％d，（（y + 1）％10））; 
} else if（y == 0）{
 printf（％d，（x％10））; 
} else if（x == y）{
 printf（\\）; 
} else if（is_perfect_square（（x * x）+（y * y）））{
 printf（*）; 
} 
} 
 printf（\\\
）; 
 
 
 
 bool is_perfect_square（int num）{
 int root =（int）（sqrt（num））; 
 
 if（num == root * root）{
 return true; 
} else {
 return false; 
} 
} 
  

仍在研究可能的解决方案。

解决方案

提示：

i，j;

  for i 0 .. limit {
 for j 0 .. limit {
 / *在这里实现下面的算法!! * / 
} 
 printf（\ n）
} 
  

要在循环中使用的算法：

• if i == 0 通过打印打印x轴值（j + 1）％10 [见注释]


• else if j == 0 通过打印打印y轴值 i％10


• else if i == j print '\'


• else if is_perfect_square（（i * i）+（j * j））返回1，print '*'
• 

• else $ print b
  
  $ b

  is_perfect_square函数的规范：函数返回1如果输入t是一个完美的正方形，否则为0。例如：
  
  

  
  
  • is_perfect_square（25） should return 1
  
  
  • is_perfect_square（7） should return 0
  
  
  • is_perfect_square（49） should return 1
  
  
  
  
  

  注意： i == 0 case should print j％10 以0开始输出以表示原点。但是提供的输出从1开始。因此使用（j + 1）％10

  
  
  

  您可能需要来处理一些角落案例，一旦这个算法在代码中实现，应该是直截了当的。

  
  

  I am trying to create a program that prints the mapping data for found pythagorean triples in C. So far, I have coded the program to be able to find the triples.

  #include <stdio.h>
  #include <math.h>
  
  int main (int argc, char * argv[]) {
      int a, b, c;
      int a2, b2, c2;
      int limit = 60;
  
      for (a = 1; a <= limit; a++) {
          a2 = a * a;
          for (b = 1; b <= limit; b++) {
              b2 = b * b;
              for (c = 0; c <= ((int)sqrt(a2+b2)); c++) {
                  c2 = c * c;
                  if (a < b && (c2 == (a2 + b2))) {
                      printf("triple: %d %d %d\n", a, b, c);
                  }
              }
          }
      }
  }
  

  The intended output is expected in the format:

  123456789012345678901234567890123456789012345678901234567890
  1\
  2 \
  3  \
  4  *\
  5    \
  6     \
  7      \
  8     * \
  9        \
  0         \
  1          \
  2    *   *  \
  

  I am trying to write a loop that does this, but cannot think of how to print in this way. Any suggestions?

  UPDATE: I managed to print the x and the y axis (x = a and y = b). The values are correct, now the mapping part is left.

      for (int x = 0; x < a; x++) { // x-axis = a
          printf("%d ", x);
      }
  
      printf("\n");
  
      for (int y = 0; y < b; y++) { // y-axis = b
  
          printf("%d\n", y);
      }
  

  UPDATE: Modified the code, the output is printing, but having problems printing spaces. tried manually adding spaces to the " \ " and " * " but that just distorts the whole image.

  #include <stdio.h>
  #include <math.h>
  #include <stdbool.h>
  
  bool is_perfect_square(int num);
  
  int main (int argc, char * argv[]) {
  
      int a, b, c;
      int a2, b2, c2;
      int limit = 60;
      bool flag = false;
  
      for (a = 1; a <= limit; a++) {
          a2 = a * a;
          for (b = 1; b <= limit; b++) {
              b2 = b * b;
              for (c = 0; c <= ((int)sqrt(a2+b2)); c++) {
                  c2 = c * c;
                  if (a < b && (c2 == (a2 + b2))) {
                      // printf("triple: %d %d %d\n", a, b, c);
                  }
              }
          }
      }
  
      for (int x = 0; x < a; x++) {
          for (int y = 0; y < b; y++) {
              if (x == 0) {
                  printf("%d ", ((y+1)% 10));
              } else if (y == 0) {
                  printf("%d ", (x % 10));
              } else if (x == y) {
                  printf("\\");
              } else if (is_perfect_square((x*x) + (y*y))) {
                  printf("*");
              }
          }
          printf("\n");
      }
  }
  
  bool is_perfect_square (int num) {
      int root = (int)(sqrt(num));
  
      if (num == root * root) {
          return true;
      } else {
          return false;
      }
  }
  

  Still working on possible solution.

  解决方案

  Hint :

  Have a nested loop with index i,j;

  for i 0.. limit {
    for j 0 .. limit {
     /*Implement the below algorithm here!!*/
    }
      printf("\n")
  }
  

  Algorithm to be used inside the loop:

  • if i == 0 print x axis values by printing (j+1)% 10 [see note at end]
  • else if j == 0 print y axis values by printing i % 10
  • else if i == j print '\'
  • else if is_perfect_square((i*i) + (j*j)) returns 1, print '*'
  • else print a space.

  Specifications for is_perfect_square function: A function that returns 1 if the input is a perfect square and 0 otherwise. For example:

  • is_perfect_square(25) should return 1
  • is_perfect_square(7) should return 0
  • is_perfect_square(49) should return 1

  Note: i == 0 case should print j%10 to start the output with 0 to represent origin. But the output provided in question starts with 1. Hence using (j+1)%10

  You might need to handle some corner cases, which should be straight forward once this algorithm is implemented in code.

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