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.
唉卡在B题1个小时……最后发现C是sb题10分钟秒了
dp:f[i][j]表示走i步到j的方案数
f[i][j]=Σf[i-1][k] | k能到j
n^2k的时间效率会T,但是发现所有的k是一个连续的区间,所以我们可以用前缀和存所有f[i-1][k]的状态,然后O(1)递推
还可以更快
注意到b把1到n的区间分成两半,而且从a开始走一定只能到达a所在的一半,所以可以再优化。期望能缩掉一半复杂度
(其实我是因为2500w状态+取模很虚所以想出这不靠谱的优化)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<deque> 9 #include<set> 10 #include<map> 11 #include<ctime> 12 #define LL long long 13 #define inf 0x7ffffff 14 #define pa pair<int,int> 15 #define pi 3.1415926535897932384626433832795028841971 16 #define mod 1000000007 17 using namespace std; 18 int n,a,b,k,L,R; 19 LL f[5010][5010]; 20 LL sum[5010],tot; 21 inline LL read() 22 { 23 LL x=0,f=1;char ch=getchar(); 24 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 25 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 26 return x*f; 27 } 28 int main() 29 { 30 n=read();a=read();b=read();k=read(); 31 if (a<b) 32 { 33 L=1;R=b-1; 34 }else 35 { 36 L=b+1;R=n; 37 } 38 f[0][a]=1; 39 for (int i=a;i<=n;i++)sum[i]=1; 40 for (int i=1;i<=k;i++) 41 { 42 for (int j=L;j<=R;j++) 43 { 44 int des=(b+j)>>1; 45 if (j<b) 46 { 47 while(b-des<=des-j) des--; 48 while(b-(des+1)>(des+1)-j) des++; 49 f[i][j]=(sum[des]-f[i-1][j]+mod)%mod; 50 }else 51 { 52 while (des-b<=j-des) des++; 53 while ((des-1)-b>j-(des-1)) des--; 54 f[i][j]=(sum[n]-sum[des-1]-f[i-1][j]+mod)%mod; 55 } 56 } 57 sum[0]=0; 58 for (int ll=1;ll<=n;ll++) 59 sum[ll]=sum[ll-1]+f[i][ll]; 60 } 61 for (int i=L;i<=R;i++) 62 tot+=f[k][i]; 63 printf("%lld\n",tot%mod); 64 }