Codeforces Round #274 (Div. 1) C. Riding in a Lift 前缀和优化dp

C. Riding in a Lift

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/480/problem/C

Description

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 Input

5 2 4 1

Sample Output

2

HINT

 

题意

有n层楼，你一开始在a层楼，实验室在b层楼，然后你可以瞎走k次

每次只要满足|now-next|<|now-b|就可以从now走到next去

然后问你一共有多少种走法

题解：

最简单就是dp[i][j]表示我现在在第i层，瞎走了j次的方案数，但是这个转移是n^3的

我们得优化一下

每次我们可以看到，从上一个状态转移过来的是一个区间，所以大概用一个线段树优化成n^2logn或者用前缀和优化成n^2的就行了

代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
#define maxn 5005
#define mod 1000000007
int dp[maxn][maxn];
int sum[maxn];
int main()
{
    int n,a,b,k;
    scanf("%d%d%d%d",&n,&a,&b,&k);
    dp[a][0]=1;
    for(int i=1;i<=n;i++)
        sum[i]=dp[i][0]+sum[i-1];
    for(int i=1;i<=k;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(j==b)continue;
            int L,R;
            if(j<b)
            {
                L = 1,R = (j+b)/2;
                if(R-j==b-R)R--;
            }
            else
            {
                L = (j+b)/2+(j+b)%2;R=n;
                if(j-L==L-b)L++;
            }
            L=max(1,L);R=min(n,R);
            dp[j][i]=((sum[R]-sum[L-1]+mod)%mod-dp[j][i-1]+mod)%mod;
        }
        for(int j=1;j<=n;j++)
        {
            sum[j]=dp[j][i]+sum[j-1];
            while(sum[j]>=mod)sum[j]-=mod;
        }
    }
    printf("%d\n",sum[n]);
}