HDUOJ 4990 Reading comprehension

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 10
3 100

Sample Output

1
5

看到n,m的范围立刻想到要用矩阵快速幂，为理解透彻，我将两种类型的矩阵都打了出来：
n为奇数时变换矩阵A：
$\left[ \begin{matrix} F_n\\ 1\\ \end{matrix} \right] =\left[ \begin{matrix} 2 &1\\ 0 & 1 \\ \end{matrix} \right] \left[ \begin{matrix} F_{n-1}\\ 1\\ \end{matrix} \right]$
n为偶数时变换矩阵B：
$\left[ \begin{matrix} F_n\\ 1\\ \end{matrix} \right] =\left[ \begin{matrix} 2 &0\\ 0 & 1 \\ \end{matrix} \right] \left[ \begin{matrix} F_{n-1}\\ 1\\ \end{matrix} \right]$
那么我们不难发现，只要把这两个矩阵乘起来 $A*B$ 就是总的变换矩阵，变换次数即为 $n/2$ ，但要注意 $n$ 为奇数时要再左乘一次 $A$ ，注意左右乘即可，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2;//矩阵的大小
ll n,mod;
struct mat
{
    ll m[N][N];
}odd,even,a,ans;

mat mul(mat a,mat b)
{
    mat tmp;
    memset(tmp.m,0,sizeof(tmp.m));
    for(int i=0;i<N;i++)
        for(int j=0;j<N;j++)
            for(int k=0;k<N;k++)
             {
                 tmp.m[i][j]=(tmp.m[i][j]%mod+a.m[i][k]*b.m[k][j]%mod)%mod;
                 if(tmp.m[i][j]<0) tmp.m[i][j]=(tmp.m[i][j]+mod)%mod;//防越界
             }
     return tmp;
}

void mat_pow(mat a,ll k)
{
    memset(ans.m,0,sizeof(ans.m));
    ans.m[0][0]=0,ans.m[1][0]=1;
    while(k>0)
    {
        if(k&1) ans=mul(a,ans);
        a=mul(a,a);
        k>>=1;
    }
}

int main(){
    while(cin>>n>>mod){
        n++;
        even.m[0][0]=2,even.m[0][1]=1,even.m[0][2]=0,even.m[0][3]=1;
        odd.m[0][0]=2,odd.m[0][1]=0,odd.m[0][2]=0,odd.m[0][3]=1;
        a=mul(even,odd);
        mat_pow(a,n/2);
        if(n%2) ans=mul(odd,ans);
        cout<<ans.m[0][0]<<endl;
    }
}

来源：CSDN

作者：旺崽

链接：https://blog.csdn.net/qq_43765333/article/details/104543055

标签

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