codeforces #274 C. Riding in a Lift dp+前缀和优化
codeforces #274 C. Riding in a Lift dp+前缀和优化
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.
dp[i][j]表示第i步到达第j层的种数。
dp[i][j]=∑ dp[i-1][l] (abs(l-j)<dp(l-b)) 即从第l层转移到第j层。
直接转移是n^3。由于abs(l-j)<dp(l-b)限制了l的范围,且范围是连续的,因此可以用前缀和维护。
#include<bits/stdc++.h> #define REP(i,a,b) for(int i=a;i<=b;i++) #define MS0(a) memset(a,0,sizeof(a)) using namespace std; typedef long long ll; const int INF=(1<<29); const int maxn=5010; const ll p=1000000007; ll n,a,b,k; ll dp[maxn][maxn]; ll sum[maxn]; int main() { freopen("in.txt","r",stdin); while(cin>>n>>a>>b>>k){ MS0(dp); for(int i=1;i<=n;i++) dp[1][i]=(abs(a-i)<abs(a-b))&&(i!=a); sum[0]=0; for(int i=1;i<=n;i++) sum[i]=(sum[i-1]+dp[1][i])%p; for(int i=2;i<=k;i++){ for(int j=1;j<=n;j++){ if(j<b){ dp[i][j]=(dp[i][j]+(sum[(b+j-1)/2]-sum[0]-(sum[j]-sum[j-1])+3*p)%p)%p; } else if(j>b){ dp[i][j]=(dp[i][j]+(sum[n]-sum[(b+j)/2]-(sum[j]-sum[j-1])+3*p)%p)%p; } } for(int j=1;j<=n;j++){ sum[j]=(sum[j-1]+dp[i][j]%p)%p; } } ll ans=0; for(int i=1;i<=n;i++){ ans=(ans%p+dp[k][i]%p)%p; } cout<<ans<<endl; } return 0; }