【2019 牛客多校第八场 Just Jump】

1.题目链接。题目大意：某人要从起点1跳到终点L，这中间有1，2，3..L这个L个点。每次跳的距离大于等于d个点。然而有些点比较特殊，这些特殊的点用二元组(ti,pi)给出，意为：在第ti步，不能落在pi这个点。这样的限制有m组，问有多少种方案可以从1安全的跳到L。

2.                      

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353ll;
const int maxn = 3000 + 5;
const int maxm = 1e7 + 7;

LL qpow(LL a, LL b) 
{
  if (b < 0)return 0;
  LL ans = 1ll;
  while (b) {
    if (b % 2)ans = ans * a % mod;
    b >>= 1;
    a = a * a % mod;
  }return ans;
}
LL fac[maxm];
LL inv_fac[maxm];
void init() 
{
  fac[0] = fac[1] = 1;
  for (int i = 2; i < maxm; i++)fac[i] = fac[i - 1] * i % mod;
  inv_fac[maxm - 1] = qpow(fac[maxm - 1], mod - 2);
  for (LL i = maxm - 2; i >= 1; i--) inv_fac[i] = (inv_fac[i + 1] * (i + 1)) % mod;
  inv_fac[0] = 1;
}

struct node {
  LL t, p;
  bool operator<(const node& a)const {
    return p < a.p;
  }
}Q[maxn];

LL L, d, m;

LL C(int n, int m) {
  if (n<0 || m>n)return 0;
  return fac[n] * inv_fac[m] % mod * inv_fac[n - m] % mod;
}
LL F(LL t, LL p) {
  if (t == 0) {
    if (p == 0)return 1;
    return 0;
  }
  if (t < 0)return 0;
  if (p < d * t)return 0;
  return C(p - d * t + t - 1, p - d * t);
}
LL dp[maxn];
LL f[maxm], pre[maxm];
int main()
{
  init();
  scanf("%lld%lld%lld", &L, &d, &m);
  for (int i = 1; i <= m; i++)
  {
    scanf("%lld%lld", &Q[i].t, &Q[i].p);
  }
  f[0] = 1;
  pre[0] = 1;
  for (int i = 1; i < d; i++)
  {
    f[i] = 0;
    pre[i] = pre[i - 1] + f[i];
  }
  for (int i = d; i <= L; i++)
  {
    f[i] = pre[i - d];
    pre[i] = (pre[i - 1] + f[i]) % mod;
  }
  sort(Q + 1, Q + m+1);
  for (int i = 1; i <= m; i++)
  {
    dp[i] = F(Q[i].t, Q[i].p);
    if (dp[i] == 0)continue;
    for (int j = 1; j <= i - 1; j++)
    {
      if (Q[j].t >= Q[i].t)continue;
      if (Q[i].p - Q[j].p < (Q[i].t - Q[j].t) * d)continue;
      dp[i] = (dp[i] - dp[j] * F(Q[i].t - Q[j].t, Q[i].p - Q[j].p)%mod + mod) % mod;
    }
  }
  LL ans = f[L];
  for (int i = 1; i <= m; i++)
  {
    ans = (ans - dp[i] * f[L - Q[i].p] % mod + mod) % mod;
  }
  cout << ans << endl;

}