hdu6395 （矩阵快速幂+分块）

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2018百度之星复赛晋级名单出炉（增加20%晋级名额）~
Sequence

Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)
Total Submission(s):     Accepted Submission(s): 

Problem Description
Let us define a sequence as below

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

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

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

≤T≤≤A,B,C,D≤≤P,n≤

Sample Input

Sample Output

Source
 Multi-University Training Contest 

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chendu   |   We have carefully selected several similar problems for you:
 

分块矩阵快速幂

构造    0 1  0            fn-2            fn-1

c d  1     *       fn-1   =       fn

0 0 1              p/i              p/i

 //这题也是利用这个性质，比如p=16,
//那么：i按照p/(p/i)+1来递增，
//得到：1，2，3，4，5，6，9，17.
//滑动的区间中每个数对于p来说除数一样，比如【9，17）这部分对于16来说除数都是1，

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
#define ll long long
const int mod =1e9+;
struct node{
    ll a[][];
};
struct pp{
    ll a[][];
};
ll num[];
ll num2[];
node juzheng(node p,node q)
{
    node ans;
    memset (ans.a,,sizeof ans.a);
    for(int i=; i<=; i++)
        for(int j=; j<=; j++)
            for(int h=; h<=; h++)
                ans.a[i][j] =(ans.a[i][j]+p.a[i][h]*q.a[h][j])%mod;

      return ans;
}
pp juzheng2(node z,pp q)
{
    pp ans;
    memset (ans.a,,sizeof ans.a);
    ans.a[][]=(z.a[][]*q.a[][]+z.a[][]*q.a[][]+z.a[][]*q.a[][])%mod;
    ans.a[][] =(z.a[][]*q.a[][]+z.a[][]*q.a[][]+z.a[][]*q.a[][])%mod;
    ans.a[][] =(z.a[][]*q.a[][]+z.a[][]*q.a[][]+z.a[][]*q.a[][])%mod;
    return ans;
}

node juzheng_qsm(node q,int m)
{
    node ans;
    memset(ans.a,,sizeof(ans.a));
    ans.a[][]=;ans.a[][]=;ans.a[][]=;
  while(m)
  {
      if(m%==)
      {
          ans=juzheng(ans,q);
      }
      m=m/;
      q=juzheng(q,q);
  }
    return ans;
}

int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        ll A,B,C,D,p,n;
        memset(num2,,sizeof num2);
        scanf("%lld%lld%lld%lld%lld%lld",&A,&B,&C,&D,&p,&n);
        if(n<=) {printf("%lld\n",n==?:B);continue;}
        ll cnt=;
        ll temp=-;

        pp S;
        S.a[][]=A;
        S.a[][]=B;
        S.a[][]=;
        node san;
        memset(san.a,,sizeof san.a);
        san.a[][]=;
        san.a[][]=C; san.a[][]=D; san.a[][]= san.a[][]=;
        ll  L=min(p,n);
        ll i;
        for( i=;i<=L;i=min(L,p/(p/i))+)
        {
          S.a[][]=p/i;
          S=juzheng2(juzheng_qsm(san,min(n,p/(p/i))+-i),S);
        }

        if(--i<n)
        {
            S.a[][]=;
           S=juzheng2(juzheng_qsm(san,n-i),S);
        }

        cout<<S.a[][]<<endl;

    }
    return ;
}