Ural-1009. K-based Numbers(简单组合数)

1009. K-based Numbers
Time limit: 1.0 second
Memory limit: 64 MB
Let’s consider K-based numbers, containing exactly N digits. We define a number to be valid if itsK-based notation doesn’t contain two successive zeros. For example:1010230 is a valid 7-digit number;
1000198 is not a valid number;
0001235 is not a 7-digit number, it is a 4-digit number.
Given two numbers N and K, you are to calculate an amount of valid K based numbers, containing Ndigits.You may assume that 2 ≤ K ≤ 10; N ≥ 2; N + K ≤ 18.
Input
The numbers N and K in decimal notation separated by the line break.
Output
The result in decimal notation.
Sample
input
output
[code]2
10
90
[/code]
题意：n位k进制的数且不能有两个以上连续的0

思路：符合要求的数之中最多含有 n/2(整除)个0。当有i个0时，非零数有(k-1)^(n-i)种情况，除了首位不能是0外，其他位都有可能为0，共有Cn(n-i,i)种情况， 其中Cn()为组合数。根据乘法原则，当有i个0时，共有(k-1)^(n-i)*Cn(n-i,i)种情况。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <algorithm>

using namespace std;
int n,k;
int power(int x,int y)
{
int res = 1;
for(int i=1;i<=y;i++)
{
res *= x;
}
return res;
}
int Cn(int x,int y)
{
int res=1;
for(int i=x;i>x-y;i--)
{
res *= i;
}
for(int i=y;i>0;i--)
{
res /= i;
}
return res;
}
void solve()
{
int ans=0,tmp1;
for(int i=0;i<=n/2;i++)
{
tmp1 = Cn(n-i,i);
ans +=(power(k-1,n-i)*tmp1);
}
printf("%d\n",ans);
}
int main()
{
//    freopen("in.txt","r",stdin);
while(scanf("%d%d",&n,&k)!=EOF)
{
solve();
}
return 0;
}