hdu 4990 Reading comprehension

Reading comprehension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 318    Accepted Submission(s): 151

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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   

看到题目，数据较大，想到可能要用到公式。

最后看了BC的题解，才知道原来是递推。

注意LL

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include<stack>
#include<queue>
#include<map>
#define PI acos(-1.0)
typedef long long LL;
const int MAX=0xfffffff;
using namespace std;
class mat{
public:
    LL v[3][3];
    mat(){
        memset(v,0,sizeof(v));
    }
};
int m;
mat matrix_mul(mat p1,mat p2)
{
    mat t ;
    for(int i=0;i<=1;i++)
        for(int j=0;j<=1;j++)
            if(p1.v[i][j])
                for(int k=0;k<=1;k++)
                    t.v[i][k]=(t.v[i][k]+p1.v[i][j]*p2.v[j][k])%m;
    return t;
}
mat matrix_mi(mat p,int k)
{
    mat t;
    for(int i=0;i<=1;i++) t.v[i][i]=1;
    while(k)
    {
        if(k&1)  t=matrix_mul(t,p);
        p=matrix_mul(p,p);
        k>>=1;
    }
    return t;
}


int main( )
{
    //freopen("1.txt","r",stdin);
    int n;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        mat a;
        a.v[0][0]=4,a.v[0][1]=1,a.v[1][1]=0,a.v[1][1]=1;
        mat ans=matrix_mi(a,n/2);
        LL aa=ans.v[0][1]*2;
        if(n&1)    aa=(aa*2+1)%m;
        else  aa=aa%m;
        printf("%I64d\n",aa);
    }
    return 0;
}