繁华模拟赛 Vicent坐电梯

Posted 2020-08-10 ACforever

/*
n<=5000
­这样就不能用O(n)的转移了，而是要用O(1)
的转移。
­注意我们每次的转移都来自一个连续的区间，
而且我们是求和
­区间求和？
­前缀和！
­令sum[step][i]表示f[step][1~i]的和
­还是以B下侧为例
­ f[step][i]=sum[step-1][i-1]+sum[step-1][k]-sum[step-1][i] 


*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
const int MAXN = 6000;
const int MOD  = 1000000007;
int f[MAXN][MAXN] = {{0}};
int n, a, b, k;

inline int getnum()
{
    char c; int ans = 0; bool flag = false;
    while ((c = getchar()) == \' \' || c == \'\\r\' || c == \'\\n\');
    if (c == \'-\') flag = true; else ans = c - \'0\';
    while ((c = getchar()) >= \'0\' && c <= \'9\') ans = ans * 10 + c - \'0\';
    return ans * (flag ? -1 : 1);
}

inline void add(int &a, int b)
{
    ll tmp = (ll)a + b;
    if (tmp >= MOD) a = (int)(tmp - MOD);
    else if (tmp < 0) a = (int)(tmp + MOD);
    else a = (int)tmp;
}

int main()
{
    freopen("lift.in", "r", stdin);
    freopen("lift.out", "w", stdout);
    n = getnum(); a = getnum(); b = getnum(); k = getnum();
    if (fabs(a - b) - 1 == 0) { printf("0\\n"); return 0; }
    for (int i = a; i <= n; i++)
        f[0][i] = 1;
    for (int step = 1; step <= k; step++)
    for (int i = 1; i <= n; i++)
        if (i == b) f[step][i] = f[step][i - 1];
        else
        {
            f[step][i] = f[step][i - 1];
            if (i < b)
            {
                int x = (b + i) / 2;
                if ((b + i) % 2 == 0) x--;
                add(f[step][i], f[step - 1][x]);
                add(f[step][i], -f[step - 1][i]);
                add(f[step][i], f[step - 1][i - 1]);
            }
            else
            {
                int x = (b + i) / 2;
                add(f[step][i], f[step - 1][n]);
                add(f[step][i], -f[step - 1][x]);
                add(f[step][i], -f[step - 1][i]);
                add(f[step][i], f[step - 1][i - 1]);
            }
        }
    printf("%d\\n", f[k][n]);
}