HDU-6395-矩阵快速幂

Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 52    Accepted Submission(s): 13

Problem Description

Let us define a sequence as below

⎧⎩⎨⎪⎪⎪⎪⎪⎪F1F2Fn===ABC⋅Fn−2+D⋅Fn−1+⌊Pn⌋

Your job is simple, for each task, you should output Fn module 109+7.

 

Input

The first line has only one integer T, indicates the number of tasks.

Then, for the next T lines, each line consists of 6 integers, A , B, C, D, P, n.

1≤T≤200≤A,B,C,D≤1091≤P,n≤109

 

Sample Input

2
3 3 2 1 3 5
3 2 2 2 1 4

 

Sample Output

36
24

 

Source

2018 Multi-University Training Contest 7

 

　　对n进行分段 ，每一段内的 floor(P/ni) 都是一定的，这样就可以对每一段直接跑矩阵快速幂了。分段的时候应该可以O(1)的，比赛时候

没想到直接上的二分= =。

　　(把第78行求边界换成这句代码就是O(1)了:          LL j=P/i==0?N:min(N,P/(P/i));

　　

 #include <bits/stdc++.h>
 using namespace std;
 #define LL long long
 LL mod=1e9+;
 LL f[],T,A,B,C,D,P,N;
 struct matrix{
     LL a[][];
     matrix(){
         memset(a,,sizeof(a));
     }
     matrix operator*(matrix &tmp){
         matrix ans;
         for(int i=;i<;++i){
             for(int j=;j<;++j){
                 for(int k=;k<;++k){
                     ans.a[i][j]=(ans.a[i][j]+a[i][k]*tmp.a[k][j]);
                 //    ans.a[i][j]%=mod;
                 }
                 ans.a[i][j]%=mod;
             }
         }
         return ans;
     }

     void show(){
         puts("---");
         for(int i=;i<;++i){
             for(int j=;j<;++j){
                 cout<<a[i][j]<<' ';
             }cout<<endl;
         }
         puts("---");
     }
 }X,U;
 matrix qpow(matrix A,int b){
     matrix ans=U;
     while(b){
         if(b&)ans=ans*A;
         A=A*A;
         b>>=;
     }
     return ans;
 }
 LL solve(LL i){
     LL key=P/i;
     LL l=i,r=N;
     while(l<r){
         int mid=r-(r-l)/;
         //cout<<mid<<' '<<P/mid<<endl;
         if(P/mid==key) l=mid;
         else if(P/mid<key){
             r=mid-;
         }
         else{
             l=mid+;
         }
     }
     return l;
 }
 int main(){
     U.a[][]=U.a[][]=U.a[][]=;
     scanf("%lld",&T);
     while(T--){
         scanf("%lld%lld%lld%lld%lld%lld",&A,&B,&C,&D,&P,&N);

         f[]=A%mod;
         f[]=B%mod;
         if(N<=){
             cout<<f[N]<<endl;
             continue;
         }
         //for(int i=3;i<=N;++i) f[i]=(C*f[i-2]%mod+D*f[i-1]%mod+(P/i))%mod;
         memset(X.a,,sizeof(X.a));
         X.a[][]=D,X.a[][]=;
         X.a[][]=C,X.a[][]=;
         LL f1=f[],f2=f[];
         for(LL i=;i<=N;){
             LL j=solve(i);
             //cout<<i<<' '<<j<<endl;
             X.a[][]=P/i;
             matrix ans=qpow(X,j-i+);
             LL _f1=(f1*ans.a[][]%mod+f2*ans.a[][]%mod+ans.a[][])%mod;
             LL _f2=(f1*ans.a[][]+f2*ans.a[][]+ans.a[][])%mod;

             f1=_f1;
             f2=_f2;
             //cout<<j<<' '<<f1<<' '<<f2<<' '<<f[j]<<' '<<f[j-1]<<endl;
             i=j+;
         }
         cout<<f1<<endl;
         //printf("%lld %lld\n",f1,f[N]);

     }
     return ;
 }
 /*
 2
 3 3 2 1 3 5
 3 2 2 2 1 4

 */