Codeforce 题目479E Riding in a Lift(DP)
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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.
a<b:
dp[i][j]=sum[i][(j+b-1)/2]-dp[i-1][j];上个位置只能在j和b的中点一下。找几个数据就得出结论了。
对于a>b:
将a,b关于对称轴对称下,那么问题就转化为a<b的问题#include<stdio.h>#include<string.h>#include<stdlib.h>#include<iostream>#include<algorithm>#define mod 1000000007using namespace std;int dp[2][5050],sum[2][5050];int n,a,b,k;void getsum(int x){ int i; for(i=1;i<=n;i++) sum[x%2][i]=(sum[x%2][i-1]+dp[x%2][i])%mod;}int main(){ //int n,a,b,k; while(scanf("%d%d%d%d",&n,&a,&b,&k)!=EOF) { if(a>b) { a=n-a+1; b=n-b+1; } memset(sum,0,sizeof(sum)); memset(dp,0,sizeof(dp)); dp[0][a]=1; getsum(0); int i,j; for(i=1;i<=k;i++) { for(j=1;j<b;j++) { dp[i%2][j]=(sum[(i-1)%2][(j+b-1)/2]-dp[(i-1)%2][j]+mod)%mod; } getsum(i); } int ans=0; for(i=1;i<=n;i++) ans=(ans+dp[k%2][i])%mod; printf("%d\n",ans); }}
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