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
题意：做电梯，刚開始的时候你在a层，不能到b层。每次你到新的地方的y，必须满足|x-y|<|x-b|，求坐k次有多少种可能
思路：比較easy想到的是dp[i][j]表示第i次到了j层的可能，分情况讨论。比如：当a<b的时候，下一次的层数i是不能超过j+(b-j-1)/2的。然后每次预先处理出前j层的可能。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int mod = 1000000007;
const int maxn = 5005;
int n, a, b, k, dp[maxn][maxn];
int sum[maxn];
int main() {
scanf("%d%d%d%d", &n, &a, &b, &k);
memset(dp, 0, sizeof(dp));
if (a < b) {
dp[0][a] = 1;
for (int j = 1; j < b; j++)
sum[j] = sum[j-1] + dp[0][j];
for (int i = 1; i <= k; i++) {
for (int j = 1; j < b; j++)
dp[i][j] = (sum[(b-j-1)/2+j] - dp[i-1][j] + mod) % mod;
sum[0] = 0;
for (int j = 1; j < b; j++)
sum[j] = (sum[j-1] + dp[i][j]) % mod;
}
printf("%d\n", sum[b-1]);
}
else {
dp[0][a] = 1;
for (int j = n; j >= b+1; j--)
sum[j] = sum[j+1] + dp[0][j];
for (int i = 1; i <= k; i++) {
for (int j = b+1; j <= n; j++)
dp[i][j] = (sum[j-(j-b-1)/2] - dp[i-1][j] + mod) % mod;
sum[0] = 0;
for (int j = n; j >= b+1; j--)
sum[j] = (sum[j+1] + dp[i][j]) % mod;
}
printf("%d\n", sum[b+1]);
}
return 0;
}
本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5261360.html，如需转载请自行联系原作者
