C 弗洛伊德'；s三角形反转

我必须制作一个按以下顺序打印的Floyd三角形：

7 8 9 10
4 5 6 
2 3
1

但目前我的代码打印如下：

1
2 3 
4 5 6
7 8 9 10

代码：

#包括
int main（）
{
int n，i，c，a=1；
scanf（“%d”和“&n”）；
对于（i=1；i我通过黑客破解解决了它（即尝试不同的东西并根据输出调整代码）：

我首先将生成代码移动到一个单独的函数，然后剪切并粘贴该函数，给它一个不同的名称，然后开始修改循环的

我浏览了四个版本，直到找到了正确的版本：

#include <stdio.h>
#include <stdlib.h>

void
fwd(int n)
{
    int i, c;
    int a = 1;

    for (i = 1; i <= n; i++) {
        for (c = 1; c <= i; c++) {
            printf("%d ", a);
            a++;
        }
        printf("\n");
    }
}

void
rev1(int n)
{
    int i, c;
    int a = 1;

    for (i = n; i >= 1; i--) {
        for (c = 1; c <= i; c++) {
            printf("%d ", a);
            a++;
        }
        printf("\n");
    }
}

void
rev2(int n)
{
    int i, c;
    int a = 0;

    for (i = 1; i <= n; i++)
        a += i;

    for (i = n; i >= 1; i--) {
        for (c = i; c >= 1; c--) {
            printf("%d ", a);
            a--;
        }
        printf("\n");
    }
}

void
rev3(int n)
{
    int i, c;
    int a = 0;

    for (i = 1; i <= n; i++)
        a += i;

    for (i = n; i >= 1; i--) {
        for (c = i; c >= 1; c--) {
            printf("%d ", (c - i) + a);
            a--;
        }
        printf("\n");
    }
}

void
rev4(int n)
{
    int i, c;
    int a = 0;

    for (i = 1; i <= n; i++)
        a += i;

    for (i = n; i >= 1; i--) {
        for (c = 1; c <= i; c++) {
            printf("%d ", (c - i) + a);
        }
        a -= i;
        printf("\n");
    }
}

int
main(int argc,char **argv)
{
    int n;

    --argc;
    ++argv;

    if (argc > 0)
        n = atoi(*argv);
    else
        scanf("%d", &n);

    printf("fwd:\n");
    fwd(n);

    printf("\nrev1:\n");
    rev1(n);

    printf("\nrev2:\n");
    rev2(n);

    printf("\nrev3:\n");
    rev3(n);

    printf("\nrev4:\n");
    rev4(n);

    return 0;
}

给你

#include <stdio.h>

int main(void) 
{
    while ( 1 )
    {
        printf( "Enter a non-negative number (0 - exit): " );
        
        unsigned int n;
        
        if ( scanf( "%u", &n ) != 1 || n == 0 ) break;
        
        int width = 0;
        
        for ( unsigned int tmp = n * ( n + 1 ) / 2; tmp != 0; tmp /= 10 )
        {
            ++width;
        }
        
        putchar( '\n' );
        
        for ( unsigned int i = 0; i < n; i++ )
        {
            unsigned int value = ( n - i ) * ( n - i + 1 ) / 2 - ( n - i - 1 );
            
            for ( unsigned int j = 0; j < n - i; j++ )
            {
                printf( "%*u ", -width, value++ );
            }
            
            putchar( '\n' );
        }
        
        putchar( '\n' );
    }
    
    return 0;
}

其他答案很好，但是没有人发布递归函数，所以我想我应该添加一个

#include <stdio.h>

void recursive(int n, int i);
void straight(int n);

// one way of doing it with a recursive function
void recursive(int n, int i) {
    if(n <= 0) 
        return;
    
    if(i <= n) {
        printf( "%-3d", (((n * (n + 1) / 2) - n) + i) );
        recursive(n, i + 1);
    } else {
        printf("\n");
        recursive(n - 1, 1);
    }
    
}

// straightforward nested loop way
void straight(int n) {
    int i, j, row_sum;
    
    for(i = n; i > 0; --i) {
        // the sum: i + (i-1) + (i-2) + ... + 2 + 1 = (i * (i+1)) / 2
        row_sum = (i * (i + 1)) / 2;
        
        for(j = i; j > 0; --j) {
            printf("%-3d", row_sum - j + 1);
        }
        printf("\n");
    }
}

// entry point
int main(int argc, char **argv) {
  int n = 0;
  
  scanf("%d", &n);
  
  printf("Recursive Output for n=%d\n", n);
  recursive(n, 1);
  
  printf("\nStraight Output for n=%d\n", n);  
  straight(n);

  return 0;
}

您可以使用后序递归函数来实现这一点。或者您可以使用基本数学来确定每行上要写入的第一个值。有一个非常常见的公式，用于计算从
1
到
x
的所有数字之和（这将给出一行中的最后一个值）。
#include <stdio.h>

int main(void) 
{
    while ( 1 )
    {
        printf( "Enter a non-negative number (0 - exit): " );
        
        unsigned int n;
        
        if ( scanf( "%u", &n ) != 1 || n == 0 ) break;
        
        int width = 0;
        
        for ( unsigned int tmp = n * ( n + 1 ) / 2; tmp != 0; tmp /= 10 )
        {
            ++width;
        }
        
        putchar( '\n' );
        
        for ( unsigned int i = 0; i < n; i++ )
        {
            unsigned int value = ( n - i ) * ( n - i + 1 ) / 2 - ( n - i - 1 );
            
            for ( unsigned int j = 0; j < n - i; j++ )
            {
                printf( "%*u ", -width, value++ );
            }
            
            putchar( '\n' );
        }
        
        putchar( '\n' );
    }
    
    return 0;
}

Enter a non-negative number (0 - exit): 10

46 47 48 49 50 51 52 53 54 55 
37 38 39 40 41 42 43 44 45 
29 30 31 32 33 34 35 36 
22 23 24 25 26 27 28 
16 17 18 19 20 21 
11 12 13 14 15 
7  8  9  10 
4  5  6  
2  3  
1  

Enter a non-negative number (0 - exit): 4

7  8  9  10 
4  5  6  
2  3  
1  

Enter a non-negative number (0 - exit): 0

#include <stdio.h>

void recursive(int n, int i);
void straight(int n);

// one way of doing it with a recursive function
void recursive(int n, int i) {
    if(n <= 0) 
        return;
    
    if(i <= n) {
        printf( "%-3d", (((n * (n + 1) / 2) - n) + i) );
        recursive(n, i + 1);
    } else {
        printf("\n");
        recursive(n - 1, 1);
    }
    
}

// straightforward nested loop way
void straight(int n) {
    int i, j, row_sum;
    
    for(i = n; i > 0; --i) {
        // the sum: i + (i-1) + (i-2) + ... + 2 + 1 = (i * (i+1)) / 2
        row_sum = (i * (i + 1)) / 2;
        
        for(j = i; j > 0; --j) {
            printf("%-3d", row_sum - j + 1);
        }
        printf("\n");
    }
}

// entry point
int main(int argc, char **argv) {
  int n = 0;
  
  scanf("%d", &n);
  
  printf("Recursive Output for n=%d\n", n);
  recursive(n, 1);
  
  printf("\nStraight Output for n=%d\n", n);  
  straight(n);

  return 0;
}

Recursive Output for n=5
11 12 13 14 15 
7  8  9  10 
4  5  6  
2  3  
1  

Straight Output for n=5
11 12 13 14 15 
7  8  9  10 
4  5  6  
2  3  
1