SGU 310. Hippopotamus

 

310. Hippopotamus

Time limit per test: 0.5 second(s)
Memory limit: 65536 kilobytes

input: standard
output: standard

After fixing your roof, you still think that it looks unpretty. So you opt for a new one, consisting of n consecutive long narrow boards. You have two types of boards: wooden ones and iron ones, giving you an amazing total of 2ⁿ possible roofs.

But the safety should not be left aside. Having considered the weight and the cruising speed of a falling hippopotamus, you decide to have at least k iron boards among every mconsecutive boards.

How many possibilities do you have?

Input

The input file contains three integers, n, m and k, separated by spaces and/or line breaks. 1 ≤ n ≤ 60, 1 ≤ m ≤ 15, 0 ≤ k ≤ m ≤ n.

Output

Output the number of possibilities.

Example(s)

sample input	sample output
10 2 1	144

sample input	sample output
5 5 2	26

sample input	sample output
3 2 2	1

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水题，简单的DP

 

 

#include <iostream>
#include<stdio.h>
using namespace std;
int a[50000];
long long b[70][50000];
int dog(int num)
{
    int tot=0;
    while(num)
    {
        tot+=(num&1);
        num=(num>>1);
    }
    return tot;
}
int main()
{
    int i,j,n,m,k;
    long long tot;
    for(i=0;i<(1<<15);i++)
        a[i]=dog(i);
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        for(i=1;i<=n;i++)
            for(j=0;j<(1<<m);j++)
                b[i][j]=0;
        for(i=0;i<(1<<m);i++)
            b[m][i]=(a[i]>=k?1:0);
        for(i=m+1;i<=n;i++)
            for(j=0;j<(1<<m);j++)
            {
                if (a[j]<k) continue;
                b[i][(j>>1)+(1<<(m-1))]+=b[i-1][j];
                b[i][j>>1]+=b[i-1][j];
            }
        tot=0;
        for(i=0;i<(1<<m);i++)
            if (a[i]>=k) tot+=b[n][i];
        printf("%lld\n",tot);

    }
    return 0;
}