#include <stdio.h>
#include <math.h>
int main(void)
{
///-------------------------------------------------------------------------------------------------------------------------------///
/// Initializes necessary variables. Description of each variable provided.
int a, b, c; // Sides of triangle
int N; // User-defined integer, where c<N
int k=0; // Counter necessary for 'if loop'
int thinA=0, thinB=0, thinC=0; // Memory for sides of 'thinnest' triangle
double totalAngle = 180; // Sum of interior angles in a triangle
///-------------------------------------------------------------------------------------------------------------------------------///
/// Introduction
printf("This program prints out all Pythagorean triples as (a,b,c) when given a positive integer, N, where c<N. \n\nThis program will also print out the number of triples and the 'thinnest' \n triangle in this range.\n\n");
///-------------------------------------------------------------------------------------------------------------------------------///
/// Requests user input for variable N. The program will then find all pythagorean triples that have side lengths less than N.
printf("Enter a positive integer: ");
scanf("%d", &N);
///-------------------------------------------------------------------------------------------------------------------------------///
/// Initilizes computing of side lengths, using several 'if' loops embedded within one another
// Side A
for (a=1; a<N; a++)
{
// Side B
for (b=1; b<N; b++)
{
// Side C
for(c=1; c<N; c++)
{
// Validation of a right angle triangle. Also validates that side A is less than side B so no triangle is listed twice
if (a*a + b*b == c*c && a < b)
{
// Prints out listed side lengths of every acceptable triangle. Also increments counter for proper print statements at end
printf("\n(%d %d %d)", a, b, c);
k++;
///-------------------------------------------------------------------------------------------------------------------------------///
/// Determination of thinnest triangle
if (atan(a*1.0/b) < totalAngle)
{
totalAngle = atan(a*1.0/b);
thinA = a;
thinB = b;
thinC = c;
}
}
}
}
}
///-------------------------------------------------------------------------------------------------------------------------------///
/// Results
// If the counter incremented (that is, a triangle was found to exist where c<N), then it will print the amount of triangles found.
// If not, it will state that no triangles were found.
if (k > 0)
{
printf("\n\nThere are %d Pythagorean triples in this range.\n", k);
printf("\nThe thinnest right-angle triangle is formed by (%d %d %d).\n\n", thinA, thinB, thinC);
}
else
printf("\nThere are no pythagorean triples.\n\n");
///-------------------------------------------------------------------------------------------------------------------------------///
/// END OF SCRIPT
///-------------------------------------------------------------------------------------------------------------------------------///
return 0;
}
晚上全部。我的代码接受用户定义的int变量N并输出范围（0，N）内的每个Pythagorean三元组。 假设我输入N为12，将打印以下内容：
Enter a positive integer: 12
(3 4 5)
(6 8 10)
There are 2 Pythagorean triples in this range.
The thinnest right-angle triangle is formed by (3 4 5).
需要做出哪些调整来制作这样的印刷顺序？
Enter a positive integer: 12
There are 2 Pythagorean triples in this range.
(3 4 5)
(6 8 10)
The thinnest right-angle triangle is formed by (3 4 5).
再次欢呼和谢谢！
答案 0 :(得分：0)
不要使用
行 printf("\n(%d %d %d)", a, b, c);
在此行中，将值存储在string类型的变量中。在这两行之后
printf("\n\nThere are %d Pythagorean triples in this range.\n", k);
printf("\nThe thinnest right-angle triangle is formed by (%d %d %d).\n\n", thinA, thinB, thinC);
添加更多一个printf，用于打印创建的字符串变量。
答案 1 :(得分：0)
我看到了几种可能性，一种最好的可能性 - 经常 - 取决于该计划的其他方面：
到目前为止，最简单的实现方法是首先计算k，打印出数字，然后重做循环以打印出结果，如下所示
// pass 1: determine k
for (a=1; a<N; a++)
{
// Side B
for (b=1; b<N; b++)
{
// Side C
for(c=1; c<N; c++)
{
// Validation of a right angle triangle. Also validates that side A is less than side B so no triangle is listed twice
if (a*a + b*b == c*c && a < b)
{
k++;
}
}
}
}
if (k > 0) {
printf("There are %d Pythagorean triples in this range.\n", k);
} else {
printf("There are no pythagorean triples.\n\n");
// we're done
return 0;
}
// pass 2 - print out the triples found and the thinnest
for (a=1; a<N; a++)
{
// Side B
for (b=1; b<N; b++)
{
// Side C
for(c=1; c<N; c++)
{
// Validation of a right angle triangle. Also validates that side A is less than side B so no triangle is listed twice
if (a*a + b*b == c*c && a < b)
{
printf("(%d %d %d)\n", a, b, c);
if (atan(a*1.0/b) < totalAngle)
{
totalAngle = atan(a*1.0/b);
thinA = a;
thinB = b;
thinC = c;
}
}
}
}
}
if (k > 0)
{
printf("The thinnest right-angle triangle is formed by (%d %d %d).\n\n", thinA, thinB, thinC);
}
这种方法的优点是不需要缓冲任何东西，没有涉及动态内存的分配，所以它很简单，但当然计算完成了两次，而且在现实生活中可能是不可接受的。另请注意，在大多数情况下，将\n放在printf的末尾会更容易，正如Basile已经指出的那样。
第一个选择是sprintf并将结果字符串strcat转换为char []变量，该变量保证足够大以包含连接结果字符串的最大长度。这样你只需执行一次计算，但随着N的增长，内存结构可能会增长到很大的比例。虽然很简单，但这种方法只适用于相当小的N值。
第三种方法是将各个结果字符串存储在链表中，每次找到结果时都分配一个节点。将它们打印出来只需要遍历链表，打印出每个节点。这是最有效和最优雅的解决方案，它避免了以前解决方案的缺点，代价是实现链表的一些额外代码。