如何在弗洛伊德三角形中找到数字属于哪一行和哪一列？
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54 55
例如，
非常感谢您！
请注意，第 n 行以 value 结尾 n*(n+1)/2。所以你可以制作二次方程并求解它以获得给定数字 k 的行号
n*(n+1)/2 = k
n^2 + n - 2*k = 0
D = 1 + 8*k
n_row = Ceil((-1 + Sqrt(D)) / 2) //round float value up
例如，对于 k=33，您可以计算
n_row = Ceil((-1 + Sqrt(265)) / 2) =
Ceil(7.639) =
8
有了n_row，求上一行的最后一个数和k在当前行中的位置
n_Column = 33 - n_row * (n_row - 1) / 2 =
33 - 28 =
5
行查找替代方法的伪代码：
sum = 0
row = 0
while sum < k do
row++
sum = sum + row
我认为这种方法更自然：
#include <iostream>
size_t getRow(size_t n)
{ // just count the rows, and when you meet the number, return the row
size_t row(0), k(1);
for (row = 1; row <= n; row++)
{
for (size_t j = 1; j <= row; j++)
{
if (k == n)
{
goto end;
}
k++;
}
}
end:return row;
}
size_t getCol(size_t n)
{ /* well, we have already found the row, so now knowing that every n-th row starts
with n(n-1)/2+1 and ends with n(n+1)/2, just count the columns and when you
meet the number (and that surely will happen), just return the column and you're done*/
size_t col(1);
size_t r = getRow(n);
for (size_t j = r * (r - 1) / 2+1; j <= r*(r+1)/2; j++)
{
if (j == n)
{
break;
}
col++;
}
return col;
}
int main()
{
size_t n;
std::cin >> n;
std::cout << "Number " << n << " lies in row " << getRow(n) << ", column " << getCol(n) << " of the Floyd's triangle.\n";
return 0;
}
在 python 中，这看起来像这样（如果你不想使用sqrt）：
def rc(n):
# rc(1) = (1, 1); rc(33) -> (8, 5)
assert n > 0 and int(n) == n
sum = 0
row = 0
while sum < n:
row += 1
sum += row
col = n - sum + row
return row, col