codeforces #274 C. Riding in a Lift dp+前缀和优化-优快云博客

本文详细解析 Codeforces #274 C 题目 Riding in a Lift 的解题思路，利用动态规划及前缀和优化算法求解在特定条件下乘坐电梯的不同路径数量，并给出具体实现代码。

codeforces #274 C. Riding in a Lift dp+前缀和优化

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

input

5 2 4 2

output

input

5 3 4 1

output

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.

dp[i][j]表示第i步到达第j层的种数。

dp[i][j]=∑ dp[i-1][l] (abs(l-j)<dp(l-b)) 即从第l层转移到第j层。

直接转移是n^3。由于abs(l-j)<dp(l-b)限制了l的范围，且范围是连续的，因此可以用前缀和维护。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
const int INF=(1<<29);
const int maxn=5010;

const ll p=1000000007;
ll n,a,b,k;
ll dp[maxn][maxn];
ll sum[maxn];

int main()
{
    freopen("in.txt","r",stdin);
    while(cin>>n>>a>>b>>k){
        MS0(dp);
        for(int i=1;i<=n;i++) dp[1][i]=(abs(a-i)<abs(a-b))&&(i!=a);
        sum[0]=0;
        for(int i=1;i<=n;i++) sum[i]=(sum[i-1]+dp[1][i])%p;
        for(int i=2;i<=k;i++){
            for(int j=1;j<=n;j++){
                if(j<b){
                    dp[i][j]=(dp[i][j]+(sum[(b+j-1)/2]-sum[0]-(sum[j]-sum[j-1])+3*p)%p)%p;
                }
                else if(j>b){
                    dp[i][j]=(dp[i][j]+(sum[n]-sum[(b+j)/2]-(sum[j]-sum[j-1])+3*p)%p)%p;
                }
            }
            for(int j=1;j<=n;j++){
                sum[j]=(sum[j-1]+dp[i][j]%p)%p;
            }
        }
        ll ans=0;
        for(int i=1;i<=n;i++){
            ans=(ans%p+dp[k][i]%p)%p;
        }
        cout<<ans<<endl;
    }
    return 0;
}