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

题意：做电梯。刚開始的时候你在a层，不能到b层，每次你到新的地方的y，必须满足|x-y|<|x-b|，求坐k次有多少种可能

思路：比較easy想到的是dp[i][j]表示第i次到了j层的可能。分情况讨论。比如：当a<b的时候。下一次的层数i是不能超过j+(b-j-1)/2的，然后每次预先处理出前j层的可能。

#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;
}