矩阵

原题http://acm.hdu.edu.cn/showproblem.php?pid=4990

本题先把方程写出来，就是把光计算n，不管m。公式为f[n] = f[n-2]*2+f[n-1]+1;

然后用矩阵快速幂。以前用的是AAAAAB型。以后都采用ABBBB型。就把这题当做模板吧。

如何计算构建矩阵：因为要从已知的是f[n-2],f[n-],我们要得到的是新矩阵里是要f[n-1],f[n]，由此来构建需要的矩阵。

因为已知的是两个数，所以只要计算n-2次就可以了

代码虐我千百遍，我待代码如初恋

Reading comprehension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 252    Accepted Submission(s): 121

Problem Description

Read the program below carefully then answer the question.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>

const int MAX=100000*2;
const int INF=1e9;

int main()
{
  int n,m,ans,i;
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    ans=0;
    for(i=1;i<=n;i++)
    {
      if(i&1)ans=(ans*2+1)%m;
      else ans=ans*2%m;
    }
    printf("%d\n",ans);
  }
  return 0;
}

 

Input

Multi test cases，each line will contain two integers n and m. Process to end of file.
[Technical Specification]
1<=n, m <= 1000000000

 

Output

For each case，output an integer，represents the output of above program.

 

Sample Input

1 103 100

 

Sample Output

15

 

#include <stdio.h>#include <stdlib.h>#include <malloc.h>#include <ctype.h>#include <limits.h>#include <string.h>#include <string>#include <math.h>#include <algorithm>#include <iostream>#include <queue>#include <stack>#include <deque>#include <vector>#include <set>#include <map>using namespace std;#define max 15struct Matrix{    __int64 m[max][max];};struct Matrix I,s;__int64 n,mod;Matrix Mul(Matrix a,Matrix b){    Matrix c;    __int64 i,j,k;    for(i=0;i<3;i++){        for(j=0;j<3;j++){            c.m[i][j] = 0;            for(k=0;k<3;k++){                c.m[i][j]+=(a.m[i][k]*b.m[k][j]);                c.m[i][j]%=mod;            }        }    }    return c;}Matrix Pow(Matrix a,__int64 n){    Matrix m,b;    m = a;    b = I;    while(n){        if(n%2){            b = Mul(b,m);        }        n/=2;        m = Mul(m,m);    }    return b;}int main(){    __int64 i,j;    __int64 k,a,b;    while(~scanf("%I64d%I64d",&n,&mod)){        Matrix ans,p,q;        memset(I.m,0,sizeof(I.m));        memset(p.m,0,sizeof(p.m));        memset(q.m,0,sizeof(q.m));        for(i=0;i<3;i++){            I.m[i][i] = 1;        }        q.m[0][0] = 1;        q.m[0][1] = 2;        q.m[0][2] = 1;        p.m[0][1] = 2;        p.m[1][0] = 1;        p.m[1][1] = 1;        p.m[2][1] = 1;        p.m[2][2] = 1;        if(n == 1){            printf("%I64d\n",1%mod);        }        else if(n == 2){            printf("%I64d\n",2%mod);        }        else{            ans = Pow(p,n-2);            ans = Mul(q,ans);            printf("%I64d\n",ans.m[0][1]%mod);        }    }        return 0;}