Sequence( 分块+矩阵快速幂 )

题目链接

复制代码

#include<bits/stdc++.h>
using namespace std;
 #define e exp(1)
 #define pi acos(-1)
 #define mod 1000000007
 #define inf 0x3f3f3f3f
 #define ll long long
 #define ull unsigned long long
 #define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b){return b?gcd(b,a%b):a;}

ll A,B,C,D,N,P;
struct mat {
    ll a[3][3];
}c;

mat mat_mul(mat x,mat y) {
    mat s;
    mem(s.a,0);
    for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)
        for(int k=0;k<3;k++)
        s.a[i][j]=(s.a[i][j]+x.a[i][k]*y.a[k][j]%mod)%mod;
    return s;
}

mat mat_pow(ll n) {
    mat res;
    mem(res.a,0);
    res.a[0][0]=res.a[1][1]=res.a[2][2]=1;
    while(n) {
        if(n&1) res=mat_mul(res,c);
        c=mat_mul(c,c);
        n>>=1;
    }
    return res;
}

ll solove(ll i) {
    ll l=i,r=N;
    ll p=P/i;
    while(l<r) {
        int mid=r-(r-l)/2;
        if(p==P/mid) l=mid;
        else if(p>P/mid)r=mid-1;
        else l=mid+1;
    }
    return l;
}

int main() {
    int T;scanf("%d",&T);
    while(T--) {
        scanf("%lld%lld%lld%lld%lld%lld",&A,&B,&C,&D,&P,&N);
        if(N==1){printf("%lld\n",A);continue;}
        if(N==2){printf("%lld\n",B);continue;}
        ll f1=B; ll f2=A; mat ans;
        for(ll i=3; i<=N;) {
            ll j=solove(i);
            mem(c.a,0);
            c.a[0][0]=D;c.a[0][1]=C;
            c.a[2][2]=1;c.a[1][0]=1;
            c.a[0][2]=P/i; ans=mat_pow(j-i+1);
            ll ff1=(ans.a[0][0]*f1%mod+ans.a[0][1]*f2%mod+ans.a[0][2])%mod;
            ll ff2=(ans.a[1][0]*f1%mod+ans.a[1][1]*f2%mod+ans.a[1][2])%mod;
            f1=ff1; f2=ff2; i=j+1;
        }
        printf("%lld\n",f1);
    }
    return 0;
}