CodeForces 479E E. Riding in a Lift

CodeForces 479E E. Riding in a Lift

题目

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).

Input

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).

Output

Print a single integer — the remainder after dividing the sought number of sequences by 1000000007 (109 + 7).

题意

一个楼有N层，你起始在A层，有个神秘实验室在B层。假设你在x层，你能走到y层，满足 |x - y| < |x - b|. 你不能走到B层或者留在原地。一共走K次问有多少种走法。

解题思路

一个很简单的题，但是有一个关键点是每次需要更新一个区间，更新区间的操作用线段树让代码很大而且复杂度高，其实可以用前缀和的方法，当需要更新区间[l, r]时，只需要更新l和r+1处（分别+val和-val），这样求前缀和就是新的dp值，总复杂度为O(N^2).

代码

#include<bits/stdc++.h>

using namespace std;

const int SIZE = 5005;
const int MOD = 1000000007;
int N, A, B, K;

int dp[SIZE][SIZE];

void slv(int x, int l, int r, int d) {
dp[x][l] += d;
dp[x][l] %= MOD;
dp[x][r] -= d;
dp[x][r] = ( dp[x][r] + MOD ) % MOD;
}

int main() {
scanf("%d %d %d %d", &N, &A, &B, &K);
dp[0][A] = 1;
for(int i = 0; i < K; i++) {
for(int j = 1; j <= N; j++) {
if(j == B) continue;
int temp = abs(j - B) - 1;
slv(i+1, max(j-temp, 1), j, dp[i][j]);
slv(i+1, j+1, min(j+temp+1, N+1), dp[i][j]);
}
int cnt = 0;
for(int j = 1; j <= N; j++) {
cnt += dp[i+1][j];
cnt %= MOD;
dp[i+1][j] = cnt;
}
}
int res = 0;
for(int i = 1; i <= N; i++) {
res += dp[K][i];
res %= MOD;
}
printf("%d\n",res);
return 0;
}