CodeForces 274E. Riding in a LiftDp

题意：从a 开始不能到达b，要坐k次电梯的满足条件 ：|x - y| < |x - b|. x是当前位置，y是目标位置 的方案数。

每次转移处理个前缀和，然后每次满足条件的和的范围是 ：1到 (y+b-1) /2

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include <math.h>
using namespace std;
typedef long long LL;

const int maxn  = 5555;
int dp[maxn][maxn];int sum[maxn][maxn];
const int mod = 1000000000+7;
int main()
{
    int n, a, b, k;
    cin >> n >> a >> b >> k;
    if (a > b) a = n + 1 - a, b = n + 1 - b;
    dp[0][a] = 1;
    for (int i = 1; i <= k; i++){
        for (int j = 1; j < b; j++)
            sum[i-1][j] = sum[i-1][j - 1] + dp[i-1][j],sum[i-1][j]%=mod;
        for (int j = 1; j < b; j++){
            int r= (j+b-1) / 2;
            dp[i][j] = sum[i - 1][r] - sum[i - 1][0] - dp[i - 1][j];
            dp[i][j] = (dp[i][j]+mod) %mod;
        }
    }
    int ans = 0;
    for (int i = 1; i < b; i++)
        ans += dp[k][i],ans%=mod;
    printf("%d\n", ans);
    return 0;
}