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題解:2018 Multi-University Training Contest 7

#include<bits/stdc++.h>
#define mul(a,b,c) (1LL*(a)*(b)%(c))
using namespace std;
typedef int ll;
const ll N=3,M=1e9+7;
struct Matrix
{
    static int n;
    ll a[N][N];
    Matrix(ll k=0)
    {
        for(int i=0; i<n; ++i)fill(a[i],a[i]+n,0),a[i][i]=k;
    }
};
int Matrix::n=N;
Matrix operator*(
const Matrix &a,const Matrix &b) { Matrix r(0); for(int i=0; i<r.n; ++i) for(int j=0; j<r.n; ++j) for(int k=0; k<r.n; ++k) r.a[i][j]=(r.a[i][j]+mul(a.a[i][k],b.a[k][j],M))%M; return r; } Matrix pow(Matrix a,ll b) { Matrix r(1); for
(; b; b>>=1,a=a*a) if(b&1)r=r*a; return r; } int main() { int t,a,b,c,d,p,n; for(scanf("%d",&t); t--;) { scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&p,&n); Matrix ans,e; ans.a[0][0]=a,ans.a[1][0]=b; e.a[1][0]=c,e.a[1][1]=d,e.a[0
][1]=e.a[1][2]=e.a[2][2]=1; for(int i=3,last; i<=n; i=last+1) { if(ans.a[2][0]=p/i)last=min(p/ans.a[2][0],n); else last=n; ans=pow(e,last-i+1)*ans; } printf("%lld\n",ans.a[1][0]); } }