E. Riding in a Lift（Codeforces Round #274）

最新推荐文章于 2020-07-23 23:12:38 发布

转载最新推荐文章于 2020-07-23 23:12:38 发布 · 152 阅读

本文探讨了一个特定的算法问题：在一栋有n层楼的大楼里，从a层出发，在不经过b层的情况下，进行k次楼层间移动的不同路径数量。通过动态规划的方法解决了这一问题，并给出了详细的实现代码。

Imagine that you are in a building that has exactly n floors. You can move between the floors in a lift. Let's number the floors from bottom to top with integers from 1 to n. Now you're on the floor number a. You are very bored, so you want to take the lift. Floor number b has a secret lab, the entry is forbidden. However, you already are in the mood and decide to make k consecutive trips in the lift.

Let us suppose that at the moment you are on the floor number x (initially, you were on floor a). For another trip between floors you choose some floor with number y (y ≠ x) and the lift travels to this floor. As you cannot visit floor b with the secret lab, you decided that the distance from the current floor x to the chosen y must be strictly less than the distance from the current floor x to floor b with the secret lab. Formally, it means that the following inequation must fulfill: |x - y| < |x - b|. After the lift successfully transports you to floor y, you write down number y in your notepad.

Your task is to find the number of distinct number sequences that you could have written in the notebook as the result of k trips in the lift. As the sought number of trips can be rather large, find the remainder after dividing the number by 1000000007 (109 + 7).

Input

The first line of the input contains four space-separated integers n, a, b, k (2 ≤ n ≤ 5000, 1 ≤ k ≤ 5000, 1 ≤ a, b ≤ n, a ≠ b).

Output

Print a single integer — the remainder after dividing the sought number of sequences by 1000000007 (109 + 7).

Sample test(s)

input
5 2 4 1

output
2

input
5 2 4 2

output
2

input
5 3 4 1

output
0

Note

Two sequences p1, p2, ..., pk and q1, q2, ..., qk are distinct, if there is such integer j (1 ≤ j ≤ k), that pj ≠ qj.

Notes to the samples:

In the first sample after the first trip you are either on floor 1, or on floor 3, because |1 - 2| < |2 - 4| and |3 - 2| < |2 - 4|.
In the second sample there are two possible sequences: (1, 2); (1, 3). You cannot choose floor 3 for the first trip because in this case no floor can be the floor for the second trip.

In the third sample there are no sought sequences, because you cannot choose the floor for the first trip.

上次的cf今天才补题o(╯□╰)o，给n层楼。在a层開始，不能在b层停，且当在x层去y层时。|x - y| < |x - b|，求运行k

次的方案数。

有两种情况，dp[i][j]，i为第i次，j为当前停的层数。

当a<b时，此时全部的x不会超过b，当第i次停在j层。第i-1次肯定在[0,(b+j-1)/2],左端点不难想到,右端点推导过程:

设第i-1次停在x层。则第i层全部大于x小于b的点都能够取。我们仅仅考虑小于x的点。则x-j<=b-x-1,

整理得: x<=(b+j-1)/2; 所以转移方程为：dp[i][j]=(sum[i-1][(j+b-1)/2]-dp[i-1][j]+mod)%mod;

当a>b时，同理得dp[i][j]=((sum[i-1][n]-sum[i-1][(j+b)/2]+mod)%mod-dp[i-1][j]+mod)%mod;

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=5000+100;
const int mod=1000000000+7;
int dp[maxn][maxn];
int sum[maxn][maxn];
int n;
void getsum(int x)
{
    for(int i=1;i<=n;i++)
    {
    sum[x][i]=(sum[x][i-1]+dp[x][i])%mod;
  //  printf("%I64d\n",sum[x][i]);
    }
}
int main()
{
    int a,b,k;
    scanf("%d%d%d%d",&n,&a,&b,&k);
    memset(dp,0,sizeof(dp));
    memset(sum,0,sizeof(sum));
    dp[0][a]=1;
    if(a<b)
    {
        getsum(0);
       for(int i=1;i<=k;i++)
       {
        for(int j=1;j<b;j++)
        {
           dp[i][j]=(sum[i-1][(j+b-1)/2]-dp[i-1][j]+mod)%mod;
          // printf("%I64d ",dp[i][j]);
        }
      // printf("\n");
        getsum(i);
       }
    }
    else
    {
        getsum(0);
        for(int i=1;i<=k;i++)
        {
            for(int j=b+1;j<=n;j++)
            {
                //printf("%d %d\n",sum[i-1])
                dp[i][j]=((sum[i-1][n]-sum[i-1][(j+b)/2]+mod)%mod-dp[i-1][j]+mod)%mod;
               // printf("%d ",dp[i][j]);
            }
            getsum(i);
        }
    }
    long long ans=0;
    for(int i=1;i<=n;i++)
    {
    ans=(ans+dp[k][i])%mod;
    //printf("%d ",dp[k][i]);
    }
    printf("%I64d\n",ans);
    return 0;
}