Codeforces Round #274 (Div. 2) E. Riding in a Lift（DP）_html/css_WEB-ITnose

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|?

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

题意：做电梯，刚开始的时候你在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：比较容易想到的是dp[i][j]表示第i次到了j层的可能，分情况讨论，例如：当a

#include #include #include #include #include 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;}