Codeforces Round #274 (Div. 2) E. Riding in a Lift（DP），

Codeforces Round #274 (Div. 2) E. Riding in a Lift（DP），

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

題意：做電梯，剛開始的時候你在a層，不能到b層，每次你到新的地方的y，必須滿足|x-y|<|x-b|，求坐k次有多少種可能

思路：比較容易想到的是dp[i][j]表示第i次到了j層的可能，分情況討論，例如：當a<b的時候，下一次的層數i是不能超過j+(b-j-1)/2的，然後每次預先處理出前j層的可能。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int mod = 1000000007;const int maxn = 5005;int n, a, b, k, dp[maxn][maxn];int sum[maxn];int main() {scanf("%d%d%d%d", &n, &a, &b, &k);memset(dp, 0, sizeof(dp));if (a < b) {dp[0][a] = 1;for (int j = 1; j < b; j++)sum[j] = sum[j-1] + dp[0][j];for (int i = 1; i <= k; i++) {for (int j = 1; j < b; j++)dp[i][j] = (sum[(b-j-1)/2+j] - dp[i-1][j] + mod) % mod;sum[0] = 0;for (int j = 1; j < b; j++)sum[j] = (sum[j-1] + dp[i][j]) % mod;}printf("%d\n", sum[b-1]);}else {dp[0][a] = 1;for (int j = n; j >= b+1; j--)sum[j] = sum[j+1] + dp[0][j];for (int i = 1; i <= k; i++) {for (int j = b+1; j <= n; j++) dp[i][j] = (sum[j-(j-b-1)/2] - dp[i-1][j] + mod) % mod;sum[0] = 0;for (int j = n; j >= b+1; j--)sum[j] = (sum[j+1] + dp[i][j]) % mod;}printf("%d\n", sum[b+1]);}return 0;}