dp训练15 codeforces 479E

令dp[k][j]表示，走k步到j的方法。我们首先预处理，走到第i个位置的时候，可以走到的地方，那么每一次，dp[k+1][a]~dp[k+1][b]（a和b是j所能到的左右端点）+=dp[k][j],可以看到这是区间更新，那么我们利用差分数组来更新即可。其中自己也是一个断点。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 5050;
ll dp[N][N];
int lf[N], rt[N];
int n, a, b, k;
const int mod = 1e9 + 7;
int main()
{
	n = read(); a = read(), b = read(), k = read();
	upd(i, 1, n)
	{
		if (i < b)
		{
			int len = b - i;
			len--;
			if (len <= 0)lf[i] = rt[i] = -1;
			else {
				lf[i] = max(1, i - len);
				rt[i] = min(n, i + len);
			}
		}
		else {
			int len = i - b;
			len--;
			if (len <= 0)lf[i] = rt[i] = -1;
			else {
				lf[i] = max(1, i - len);
				rt[i] = min(n, i + len);
			}
		}
	}
	if (lf[a] == -1)
	{
		cout << 0 << endl; return 0;
	}
	upd(i, lf[a], rt[a]) { if (i == a)continue; dp[1][i] = 1; }
	upd(i, 2, k)
	{
		upd(j, 1, n)
		{
			if (lf[j] == -1)continue;
			dp[i][lf[j]] = (dp[i][lf[j]] + dp[i - 1][j]) % mod;
			dp[i][j] = (dp[i][j] - dp[i - 1][j] + mod) % mod;
			dp[i][j + 1] = (dp[i][j + 1] + dp[i - 1][j]) % mod;
			dp[i][rt[j] + 1] = (dp[i][rt[j] + 1] - dp[i - 1][j] + mod) % mod;
		}
		upd(j, 1, n)
		{
			dp[i][j] = (dp[i][j] + dp[i][j - 1]) % mod;
		}
	}
	ll ans = 0;
	upd(j, 1, n)ans = (ans + dp[k][j]) % mod;
	cout << ans << endl;
	return 0;
}