如何在Floyd Triangle中找到数字所属的行和列?
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,
提前非常感谢你!
答案 0 :(得分:2)
请注意第n行ends with value n*(n+1)/2
。所以你可以制作二次方程并求解它以得到给定数k的行数
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n_row,找到当前行中k的前一行的最后一个行和k的位置
n_Column = 33 - n_row * (n_row - 1) / 2 =
33 - 28 =
5
用于行寻找的替代方法的伪代码:
sum = 0
row = 0
while sum < k do
row++
sum = sum + row
答案 1 :(得分:0)
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;
}
答案 2 :(得分:0)
在 python 中,这看起来像这样(如果您不想使用 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