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最新推荐文章于 2022-07-16 18:47:50 发布

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最新推荐文章于 2022-07-16 18:47:50 发布

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文章标签：矩阵的快速幂

本文链接：https://blog.youkuaiyun.com/baidu_30309461/article/details/47682083

Gauss Fibonacci

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2848 Accepted Submission(s): 1178

Problem Description

Without expecting, Angel replied quickly.She says: "I'v heard that you'r a very clever boy. So if you wanna me be your GF, you should solve the problem called GF~. "
How good an opportunity that Gardon can not give up! The "Problem GF" told by Angel is actually "Gauss Fibonacci".
As we know ,Gauss is the famous mathematician who worked out the sum from 1 to 100 very quickly, and Fibonacci is the crazy man who invented some numbers.

Arithmetic progression:
g(i)=k*i+b;
We assume k and b are both non-nagetive integers.

Fibonacci Numbers:
f(0)=0
f(1)=1
f(n)=f(n-1)+f(n-2) (n>=2)

The Gauss Fibonacci problem is described as follows:
Given k,b,n ,calculate the sum of every f(g(i)) for 0<=i<n
The answer may be very large, so you should divide this answer by M and just output the remainder instead.

Input

The input contains serveral lines. For each line there are four non-nagetive integers: k,b,n,M
Each of them will not exceed 1,000,000,000.

Output

For each line input, out the value described above.

Sample Input

  
   2 1 4 100
2 0 4 100

Sample Output

  
   21
12

Author

DYGG

Source

HDU “Valentines Day” Open Programming Contest 2007-02-14

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矩阵构造求和，然后在求模，矩阵的快速幂。

#include <iostream>
using namespace std;
const int MAX=4;
const int maxl=2;
int Mod;
typedef struct
{
 long long m[MAX][MAX];
}Matrix;
typedef struct
{
 long long m[maxl][maxl];
}Matrix2;
Matrix P={0,0,1,0,
          0,0,0,1,
          0,0,1,0,
          0,0,0,1};
Matrix I={1,0,0,0,
          0,1,0,0,
          0,0,1,0,
          0,0,0,1};
Matrix2 P1={0,1,
           1,1};
Matrix2 I1={1,0,
           0,1};
Matrix matrixmul(Matrix a,Matrix b)
{
 Matrix c;
 for(int i=0;i<MAX;i++)
  for(int j=0;j<MAX;j++)
  {
   c.m[i][j]=0;
   for(int k=0;k<MAX;k++)
   {
    c.m[i][j]+=(a.m[i][k]*b.m[k][j])%Mod;
   }
   c.m[i][j]=c.m[i][j]%Mod;
  }
 return c;
}
Matrix quickpow(long long n)
{
 Matrix m=P,b=I;
 while(n>=1)
 {
  if(n&1)
  b=matrixmul(b,m);
  n=n>>1;
  m=matrixmul(m,m);
 }
 return b;
}
Matrix2 matrixmull(Matrix2 a,Matrix2 b)
{
 Matrix2 c;
 for(int i=0;i<maxl;i++)
  for(int j=0;j<maxl;j++)
  {
   c.m[i][j]=0;
   for(int k=0;k<maxl;k++)
   {
    c.m[i][j]+=(a.m[i][k]*b.m[k][j])%Mod;
   }
   c.m[i][j]=c.m[i][j]%Mod;
  }
 return c;
}
Matrix2 quickpowl(long long n)
{
 Matrix2 m=P1,b=I1;
 while(n>=1)
 {
  if(n&1)
  b=matrixmull(b,m);
  n=n>>1;
  m=matrixmull(m,m);
 }
 return b;
}
int main()
{
    Matrix2 tp1,tp2;
    Matrix tp;
    long long k,b,n,m,sum;
    while(cin>>k>>b>>n>>Mod)
    {
     tp1=quickpowl(k);
     tp2=quickpowl(b);
     P.m[0][0]=tp1.m[0][0];
     P.m[0][1]=tp1.m[0][1];
     P.m[1][0]=tp1.m[1][0];
     P.m[1][1]=tp1.m[1][1];
     tp=quickpow(n);
     sum=((tp2.m[0][0]%Mod*tp.m[0][3]%Mod)+(tp2.m[0][1]%Mod*tp.m[1][3]%Mod))%Mod;
     cout<<sum<<endl;
    }
    return 0;
}