比较综合的一道题,需要用到中国剩余定理来进行取模.
Code:
#include <cstdio>
#include <algorithm>
#define N 1000006
#define ll long long
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
ll F[N];
int array[10]={0,3,5,6793,10007};
struct Node {
ll x,y;
Node(ll x=0,ll y=0):x(x),y(y){}
}arr[N];
bool cmp(Node a,Node b) {
return a.x==b.x?a.y<b.y:a.x<b.x;
}
ll qpow(ll base,ll k,ll mod) {
ll tmp=1;
for(;k;k>>=1,base=base*base%mod)
if(k&1) tmp=tmp*base%mod;
return tmp;
}
struct Lucas {
int mod;
int fac[N];
int inv(int i) {
return (int)qpow(i,mod-2,mod);
}
void init(int p) {
mod=p,fac[0]=1;
for(int i=1;i<=mod;++i) fac[i]=(ll)fac[i-1]*i%mod;
}
int C(int x,int y) {
if(y>x) return 0;
if(y==0) return 1;
return (int)(1ll*fac[x]*inv(fac[y])%mod*inv(fac[x-y])%mod);
}
int solve(ll x,ll y) {
if(y>x) return 0;
if(y==0) return 1;
return (int)(1ll*solve(x/mod,y/mod)*C(x%mod,y%mod)%mod);
}
}comb[8];
struct excrt {
ll arr[N],brr[N];
ll exgcd(ll a,ll b,ll &x,ll &y) {
if(!b) {
x=1,y=0;
return a;
}
ll gcd=exgcd(b,a%b,x,y),tmp=x;
x=y,y=tmp-a/b*y;
return gcd;
}
ll Excrt() {
int i,j;
ll ans=arr[1],M=brr[1];
for(i=2;i<=4;++i) {
ll a=M,b=brr[i],c=arr[i]-ans,gcd,x,y;
gcd=exgcd(a,b,x,y),b=abs(b/gcd);
x=(x*(c/gcd)%b+b)%b;
ans+=M*x;
M*=brr[i]/__gcd(brr[i],M);
ans=(ans%M+M)%M;
}
return ans;
}
}crt;
ll C(ll a,ll b,int ty) {
if(ty==0)
return comb[0].solve(a,b);
else {
int i,j;
for(i=1;i<=4;++i) {
crt.arr[i]=comb[i].solve(a,b);
crt.brr[i]=array[i];
}
}
return crt.Excrt();
}
int main() {
int i,j,k,flag;
// setIO("input");
ll n,m,mod;
scanf("%lld%lld%d%lld",&n,&m,&k,&mod),flag=(mod==1019663265);
if(!flag) {
comb[0].init(mod);
}
else {
for(i=1;i<=4;++i)
comb[i].init(array[i]);
}
for(i=1;i<=k;++i)
scanf("%lld%lld",&arr[i].x,&arr[i].y);
arr[++k].x=n,arr[k].y=m;
sort(arr+1,arr+1+k,cmp);
for(i=1;i<=k;++i) {
F[i]=C(arr[i].x+arr[i].y,arr[i].y,flag);
for(j=1;j<i;++j) {
if(arr[j].y<=arr[i].y)
F[i]=(F[i]-(F[j]*C(arr[i].x-arr[j].x+arr[i].y-arr[j].y,arr[i].y-arr[j].y,flag)%mod)+mod)%mod;
}
}
printf("%lld\n",F[k]);
return 0;
}
BZOJ 3782: 上学路 Lucas+ExCRT+容斥+dp
原文:https://www.cnblogs.com/guangheli/p/11514720.html