Then:
f( 1) = 1
f(2) = 2
Right n>2, there are:
F(n) = (number of methods with one grid first and then n- 1 grid)+(number of methods with two grids first and then n-2 grid)
that is
f(n) = f(n- 1) + f(n-2)
So f(n) is the n+ 1 term of Fibonacci sequence?
# include & ltstdio.h & gt
Long fiber (integer)
{if (n == 1 || n == 2) returns1;
Return fiber (n-1)+fiber (n-2);
}
Master ()
{ int n；
scanf("%d "，& ampn)；
printf("%ld\n "，fib(n+ 1))；
Returns 0;
}
/*
Exercise 9: Go up the stairs ★★
To go up the stairs, you can go up the stairs one by one, or you can come two by two.
You can also take a step or two, or even a k-step.
The problem is that if the number of last steps is n, each step can go up to 1 to k,
How many roads did you walk?
If n = 5 and k = 2, yes.
1, 1, 1, 1, 1
2, 1, 1, 1
1,2, 1, 1
1, 1,2, 1
1, 1, 1,2
2,2, 1
2, 1,2
1,2,2
There are eight kinds.
Enter the number of steps n (1
5 2
5 3
Output:
eight
13
Difficulty: easy
*/
# include & ltstdio.h & gt
# include & ltstring.h & gt
int d[3 1][3 1]；
int dfs(int left，int k)
{
int I；
if(d[left][k]！ =- 1) returns d [left] [k];
If (! Left) Return to1;
d[left][k]= 0；
for(I = 1； I < = ((left <; k)？ Left: k); i++)
d[left][k]+=dfs(left-i，k)；
Return d[ left] [k];
}
int main(void)
{
int n，k；
memset(d，- 1，sizeof(d))；
while(scanf("%d%d "，& ampn & amp； k)！ =EOF)
printf("%d\n "，dfs(n，k))；
Returns 0;
}