cf479E Riding in a Lift
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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)
#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>#include<queue>#include<deque>#include<set>#include<map>#include<ctime>#define LL long long#define inf 0x7ffffff#define pa pair<int,int>#define pi 3.1415926535897932384626433832795028841971#define mod 1000000007using namespace std;int n,a,b,k,L,R;LL f[5010][5010];LL sum[5010],tot;inline LL read(){ LL x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}int main(){n=read();a=read();b=read();k=read();if (a<b){L=1;R=b-1;}else {L=b+1;R=n;}f[0][a]=1;for (int i=a;i<=n;i++)sum[i]=1;for (int i=1;i<=k;i++){for (int j=L;j<=R;j++) { int des=(b+j)>>1; if (j<b) { while(b-des<=des-j) des--;while(b-(des+1)>(des+1)-j) des++;f[i][j]=(sum[des]-f[i-1][j]+mod)%mod; }else { while (des-b<=j-des) des++; while ((des-1)-b>j-(des-1)) des--; f[i][j]=(sum[n]-sum[des-1]-f[i-1][j]+mod)%mod; } }sum[0]=0;for (int ll=1;ll<=n;ll++) sum[ll]=sum[ll-1]+f[i][ll];}for (int i=L;i<=R;i++) tot+=f[k][i];printf("%lld\n",tot%mod);}
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