1942. Paths on a
Grid
Imagine you are attending your math lesson at school.
Once again, you are bored because your teacher tells things that you already
mastered years ago (this time he's explaining that (a + b)2 = a2 + 2ab + b2). So you decide to waste your time with drawing
modern art instead.
Fortunately you have a piece of squared paper and you
choose a rectangle of size n * m on the
paper. Let's call this rectangle together with the lines it contains a grid.
Starting at the lower left corner of the grid, you move your pencil to the
upper right corner, taking care that it stays on the lines and moves only to
the right or up. The result is shown on the left:
Really a
masterpiece, isn't it? Repeating the procedure one more time, you arrive with
the picture shown on the right. Now you wonder: how many different works of art
can you produce?
Input. The input contains several testcases. Each is
specified by two unsigned 32-bit integers n
and m, denoting the size of the
rectangle. As you can observe, the number of lines of the corresponding grid is
one more in each dimension. Input is terminated by n = m = 0.
Output. For each test case output on a line the number of
different art works that can be generated using the procedure described above.
That is, how many paths are there on a grid where each step of the path
consists of moving one unit to the right or one unit up? You may safely assume
that this number fits into a 32-bit unsigned integer.
Sample Input
5 4
1 1
0 0
Sample
Output
126
2
РЕШЕНИЕ
геометрия
Анализ алгоритма
Искомый путь – ломанная,
состоящая из n + m звеньев. n звеньев из них должны идти вертикально, остальные – горизонтально. Количество
вариантов выбрать n вертикальных звеньев из n + m
равно 
.
Реализация алгоритма
#include <stdio.h>
#include <math.h>
unsigned long
long gcd(unsigned
long long a, unsigned long long b)
{
return (!b) ? a : gcd(b,a % b);
}
unsigned long
long Cnk(unsigned
long long n, unsigned long long k)
{
unsigned long long _CnkRes = 1, t, i, i1;
if (k > n - k) k = n - k;
for(i = 1; i <= k; i++)
{
t = gcd(_CnkRes,
i);
_CnkRes = (_CnkRes
/ t) * ((n - i + 1) / (i / t));
}
return _CnkRes;
}
unsigned long
long n, k, res;
int main(void)
{
while(scanf("%llu
%llu",&n,&k), n + k)
{
res = Cnk(n + k,k);
printf("%llu\n",res);
}
return 0;
}