cf479E Riding in a Lift

E. Riding in a Lift

time limit per test 2 seconds

memory limit per test 256 megabytes

input standard input

output standard output

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:

1. 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|.
2. 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.
3. In the third sample there are no sought sequences, because you cannot choose the floor for the first trip.

 

 

唉卡在B题1个小时……最后发现C是sb题10分钟秒了

dp：f[i][j]表示走i步到j的方案数

f[i][j]=Σf[i-1][k] | k能到j

n^2k的时间效率会T，但是发现所有的k是一个连续的区间，所以我们可以用前缀和存所有f[i-1][k]的状态，然后O(1)递推

还可以更快

注意到b把1到n的区间分成两半，而且从a开始走一定只能到达a所在的一半，所以可以再优化。期望能缩掉一半复杂度

（其实我是因为2500w状态+取模很虚所以想出这不靠谱的优化）

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
#define mod 1000000007
using namespace std;
int n,a,b,k,L,R;
LL f[5010][5010];
LL sum[5010],tot;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int main()
{
    n=read();a=read();b=read();k=read();
    if (a<b)
    {
        L=1;R=b-1;
    }else 
    {
        L=b+1;R=n;
    }
    f[0][a]=1;
    for (int i=a;i<=n;i++)sum[i]=1;
    for (int i=1;i<=k;i++)
    {
        for (int j=L;j<=R;j++)
          {
              int des=(b+j)>>1;
              if (j<b)
              {
                  while(b-des<=des-j) des--;
                while(b-(des+1)>(des+1)-j) des++;
                f[i][j]=(sum[des]-f[i-1][j]+mod)%mod;
              }else
              { 
                  while (des-b<=j-des) des++;
                  while ((des-1)-b>j-(des-1)) des--;
                  f[i][j]=(sum[n]-sum[des-1]-f[i-1][j]+mod)%mod;
              }
          }
        sum[0]=0;
        for (int ll=1;ll<=n;ll++)
          sum[ll]=sum[ll-1]+f[i][ll];
    }
    for (int i=L;i<=R;i++)
      tot+=f[k][i];
    printf("%lld\n",tot%mod);
}