I-Just Jump_2019牛客暑期多校训练营(第八场)
题目链接
题意
有L+1个点,初始在第0个点上,要跳到第L个点,每次至少跳d格,也就是在点x至少要跳到x+d,且有m个限制
\((t_i, p_i)\)指跳第\(t_i\)次不能跳到\(p_i\)上
题解
设dp[i]表示从0跳到i且没有限制的方案数,可以预处理。对限制按t从小到大排序,\(g[i][0]\)表示通过了前i-1个限制,踩在了第i个限制上且1-i这些限制踩了偶数个,\(g[i][1]\)表示踩了奇数个限制的方案数,可以\(m^2\)递推,根据容斥原理加上踩偶次的方案数扣掉踩奇数次的方案数。
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mx = 1e7+5;
const int mod = 998244353;
ll dp[mx], fac[mx], invf[mx];
int g[3005][2];
int L, d, m;
int pow_mod(ll a, ll b) {
ll ans = 1;
while (b > 0) {
if (b & 1) ans = ans * a % mod;
a = a * a % mod;
b /= 2;
}
return ans;
}
struct Node {
int t, p;
bool operator < (Node other) const {
return t < other.t;
}
}node[3005];
int C(int x, int y) {
if (x < y || y < 0 || x < 0) return 0;
return (1LL * fac[x] * invf[y] % mod * invf[x-y]) % mod;
}
int calc(int i, int j) {
if (1LL * (node[j].t-node[i].t)*d > node[j].p-node[i].p) return 0;
return C(node[j].p-node[i].p-(node[j].t-node[i].t)*d + (node[j].t-node[i].t) - 1, node[j].t-node[i].t-1);
}
int main() {
scanf("%d%d%d", &L, &d, &m);
for (int i = 1; i <= m; i++) scanf("%d%d", &node[i].t, &node[i].p);
sort(node+1, node+1+m);
node[0].t = node[0].p = 0;
dp[0] = 1;
int sum = 0;
for (int i = 1; i <= L; i++) {
if (i-d >= 0) sum = (sum + dp[i-d]) % mod;
dp[i] = sum;
}
fac[0] = invf[0] = 1;
for (int i = 1; i <= L; i++) fac[i] = 1LL * fac[i-1] * i % mod;
invf[L] = pow_mod(fac[L], mod-2);
for (int i = L-1; i >= 1; i--) invf[i] = 1LL * invf[i+1] * (i+1) % mod;
g[0][0] = 1; g[0][1] = 0;
for (int i = 1; i <= m; i++) {
for (int j = 0; j < i; j++) {
g[i][0] = (g[i][0] + 1LL * g[j][1] * calc(j, i) % mod) % mod;
g[i][1] = (g[i][1] + 1LL * g[j][0] * calc(j, i) % mod) % mod;
}
}
ll ans = dp[L];
for (int i = 1; i <= m; i++) {
ans += 1LL * (g[i][0] - g[i][1]) * dp[L-node[i].p] % mod;
ans = (ans % mod + mod) % mod;
}
printf("%lld\n", ans);
return 0;
}