2 seconds
256 megabytes
standard input
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).
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).
Print a single integer — the remainder after dividing the sought number of sequences by 1000000007 (109 + 7).
5 2 4 1
2
5 2 4 2
2
5 3 4 1
0
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:
- 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|.
- 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.
- In the third sample there are no sought sequences, because you cannot choose the floor for the first trip.
【题意】有一n层楼的楼房，可坐电梯上下。初始位置在a层，b层楼门无法打开，所以无法到达。如果你当前在x层，你能走到y层当且仅当|x - y| < |x - b|.
每一次有效的移动可到达一个楼层，然后把楼层号写下，连续的移动就可写下一个序列。
问经过k次连续的移动后，产生的序列种数。
【分析】DP。dp[i][j]表示第j次移动到达i层楼的序列数，dp[i][j]=∑(dp[能够到达i层楼的楼层][j-1])%mod。所以这里需要求一个前缀和，然后去掉dp[i][j-1]。
#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
using namespace std;
typedef long long LL;
const int N = 5e3+;
const int mod = 1e9+;
int n,a,b,k;
LL dp[N][N],sum[N];
int main(){
scanf("%d%d%d%d",&n,&a,&b,&k);
dp[a][]=;
for(int i=;i<=n;i++){
sum[i]=(sum[i-]+dp[i][])%mod;
}
for(int j=;j<=k;j++){
for(int i=;i<=n;i++){
if(i>b){
int low=(i+b)/;
int up=n;
dp[i][j]=((sum[up]-sum[low]+mod)%mod-dp[i][j-]+mod)%mod;
}
else if(i<b){
int low=;
int up=(i+b)&==?(i+b)/:(i+b)/-;;
dp[i][j]=((sum[up]-sum[low]+mod)%mod-dp[i][j-]+mod)%mod;
}
//printf("i:%d j:%d dp:%lld\n",i,j,dp[i][j]);
}
sum[]=;
for(int i=;i<=n;i++){
sum[i]=(sum[i-]+dp[i][j])%mod;
}
}
LL ans=;
for(int i=;i<=n;i++)ans=(ans+dp[i][k])%mod;
printf("%lld\n",ans);
return ;
}