BZOJ3782 上学路线 【dp + Lucas + CRT】

题目链接

BZOJ3782

题解

我们把终点也加入障碍点中，将点排序，令\(f[i]\)表示从\((0,0)\)出发，不经过其它障碍，直接到达\((x_i,y_i)\)的方案数
首先我们有个大致的方案数\({x_i + y_i \choose x_i}\)
但是中途可能会经过一些其它障碍点，那么就减去
所以

\[f[i] = {x_i + y_i \choose x_i} - \sum\limits_{j = 1}^{i - 1} {x_i - x_j + y_i - y_j \choose x_i - x_j}f[j] \]

由于坐标很大，又观察到一种模数不大，一种模数为合数，且最大质因子也不大
所以可以\(Lucas\)定理 + CRT合并

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 205,maxm = 1000005,INF = 1000000000;
inline LL read(){
	LL out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
LL N,M,T,P;
struct point{LL x,y;}p[maxn];
inline bool operator <(const point& a,const point& b){
	return a.x == b.x ? a.y < b.y : a.x < b.x;
}
int pr[10],pi;
LL fac[5][maxm],fv[5][maxm],inv[5][maxm];
void sp(){
	int x = P;
	for (int i = 2; i * i <= x; i++)
		if (x % i == 0){
			pr[++pi] = i;
			x /= i;
		}
	if (x - 1) pr[++pi] = x;
}
void init(){
	for (int j = 1; j <= pi; j++){
		fac[j][0] = fac[j][1] = fv[j][0] = fv[j][1] = inv[j][0] = inv[j][1] = 1;
		int p = pr[j];
		for (int i = 2; i < pr[j]; i++){
			fac[j][i] = 1ll * fac[j][i - 1] * i % p;
			inv[j][i] = 1ll * (p - p / i) * inv[j][p % i] % p;
			fv[j][i] = 1ll * fv[j][i - 1] * inv[j][i] % p;
		}
	}
}
LL Lucas(LL n,LL m,int p){
	if (m > n) return 0;
	if (n < pr[p] && m < pr[p])
		return 1ll * fac[p][n] * fv[p][m] % pr[p] * fv[p][n - m] % pr[p];
	return 1ll * Lucas(n % pr[p],m % pr[p],p) * Lucas(n / pr[p],m / pr[p],p) % pr[p];
}
LL C(LL n,LL m){
	if (m > n) return 0;
	LL re = 0;
	for (int i = 1; i <= pi; i++){
		re = (re + 1ll * Lucas(n,m,i) * (P / pr[i]) % P * inv[i][P / pr[i] % pr[i]] % P) % P;
	}
	return re;
}
LL f[maxn];
int main(){
	N = read(); M = read(); T = read(); P = read();
	sp(); init();  //REP(i,pi) printf("%d ",pr[i]); puts("");
	REP(i,T) p[i].x = read(),p[i].y = read();
	++T;
	p[T].x = N,p[T].y = M;
	sort(p + 1,p + T + 1);
	for (int i = 1; i <= T; i++){
		f[i] = C(p[i].x + p[i].y,p[i].x);
		for (int j = 1; j < i; j++)
			if (p[j].x <= p[i].x && p[j].y <= p[i].y)
				f[i] = (f[i] - 1ll * f[j] * C(p[i].x - p[j].x + p[i].y - p[j].y,p[i].x - p[j].x) % P) % P;
		f[i] = (f[i] + P) % P;
		if (p[i].x == N && p[i].y == M){
			printf("%lld\n",f[i]);
		}
	}
	return 0;
}