hdu 4990 Reading comprehension（矩阵乘法）

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

根据上面的代码可以发现就是让求一个数组的第n项

奇数项等于前一项*2+1，偶数项等于前一项*2：

a0 = 0；

a1 = a0*2+1;

a2 = a1*2；

a3 = a2*2+1;

a4 = a3*2;

如果直接构造矩阵求第n项好像不太好求，但是我们发现偶数项或者奇数项之间的公式是固定的，我们就是偶数项为例吧。

可以推出：

a2 = 4*a0 + 2;

a4 = 4*a2 + 2;

这样就可以很轻松的构造出矩阵，如果n是偶数的话，直接就可以求出答案，如果是奇数项，那么就*2+1.

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define LL long longusing namespace std;int mod;struct Matrix{    long long m[2][2];    int n;    Matrix(int x)    {        n = x;        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)                m[i][j] = 0;    }    Matrix(int _n,int a[2][2])    {        n = _n;        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)            {                m[i][j] = a[i][j];            }    }};Matrix operator *(Matrix a,Matrix b){    int n = a.n;    Matrix ans = Matrix(n);    for(int i=0;i<n;i++)        for(int j=0;j<n;j++)            for(int k=0;k<n;k++)            {                ans.m[i][j] += (a.m[i][k]%mod)*(b.m[k][j]%mod)%mod;                ans.m[i][j] %= mod;            }    return ans;}Matrix operator ^(Matrix a,int k){    int n = a.n;    Matrix c(n);    int i,j;    for(i=0;i<n;i++)        for(j=0;j<n;j++)            c.m[i][j] = (i==j);    for(;k;k>>=1)    {        if(k&1)            c=c*a;        a = a*a;    }    return c;}int main(void){    int n,m,i,j;    while(scanf("%d%d",&n,&m)==2)    {        mod = m;        int a[2][2] = { 4,0,                        2,1};        Matrix A(2,a);        A = A^(n/2);        LL ans = A.m[1][0];        if(n % 2 == 1)            ans = (ans*2+1)%mod;        cout << ans << endl;    }    return 0;}