【JZOJ 4709】Matrix

Description

Solution

每个点往下走*b，往右走*a，走到终点的方案数几对答案的贡献，
所以对于每个已知的点，直接计算对答案的影响即可
复杂度：$O(n)$

Code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long LL;
const int N=100500,mo=1000000007;
LL read(LL &n)
{
    char ch=getchar();
    while((ch!='-')&&((ch<'0')||(ch>'9')))ch=getchar();
    LL q=0,w=1;if(ch=='-') w=-1;else q=q*10+(LL)(ch)-48;ch=getchar();
    while(ch>='0' && ch<='9')q=q*10+(LL)(ch)-48,ch=getchar();n=q*w;return(n);
}
LL m,n,ans,a,b,an,bn;
LL jc[N*2],jcn[N*2];
LL ksm(LL q,LL e)
{
    LL ans=1;
    while(e)
    {
        if(e&1)ans=(ans*q)%mo;
        e>>=1;q=(q*q)%mo;
    }
    return ans;
}
LL C(int q,int w){return jc[w]*jcn[q]%mo*jcn[w-q]%mo;}
int main()
{
    LL q,w,e;
    read(n);read(a),read(b);
    if(!a)
    {
        fo(i,1,2*n)read(q);
        printf("%lld\n",q*ksm(b,n-1)%mo);return 0;
    }
    if(!b)
    {
        fo(i,1,n)read(q);
        printf("%lld\n",q*ksm(a,n-1)%mo);return 0;
    }
    an=ksm(a,mo-2);bn=ksm(b,mo-2);
    jc[0]=jcn[0]=1;fo(i,1,2*n)jc[i]=i*jc[i-1]%mo,jcn[i]=ksm(jc[i],mo-2);
    read(e);
    q=ksm(a,n-1);w=ksm(b,n-2);
    fo(i,2,n)
    {
        read(e);
        ans=(ans+q*w%mo*e%mo*C(n-i,n*2-i-2)%mo)%mo;
        w=w*bn%mo;
    }
    read(e);
    q=ksm(a,n-2);w=ksm(b,n-1);
    fo(i,2,n)
    {
        read(e);
        ans=(ans+q*w%mo*e%mo*C(n-i,n*2-i-2)%mo)%mo;
        q=q*an%mo;
    }
    printf("%lld\n",ans);
    return 0;
}