Reading comprehension HDU

Reading comprehension HDU - 4990

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

       1

       5

   题意：就是上面的那个程序， 数据范围很大

   

   思路： 妥妥的矩阵快速幂， Fn = Fn-1 + 2Fn-2 + 1;

   构造矩阵然后用模板就行。

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<string>#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<queue>#include<stack>#include<vector>#include<set>#include<algorithm>#define maxn 10010#define INF 0x3f3f3f3f#define eps 1e-8#define MOD 1000000007#define ll long longusing namespace std;ll mod;struct Node{    ll mat[3][3];};Node mul(Node a,Node b){    Node ret;    memset(ret.mat,0,sizeof ret.mat);    for(int i=0;i<3;++i)        for(int j=0;j<3;++j)        {            ret.mat[i][j]=0;            for(int k=0;k<3;++k)                ret.mat[i][j]=(ret.mat[i][j]+a.mat[i][k]*b.mat[k][j]%mod+mod)%mod;        }    return ret;}Node Pow(Node a,ll n){    Node ret;    memset(ret.mat,0,sizeof(ret.mat));    for(int i=0;i<3;++i)    ret.mat[i][i]=1;    Node tmp=a;    while(n)    {        if(n&1)ret=mul(ret,tmp);        tmp=mul(tmp,tmp);        n>>=1;    }    return ret;}int main(){     ll n,m;     Node A;     memset(A.mat,0,sizeof A.mat);     A.mat[0][0]=A.mat[0][2]=A.mat[1][0]=A.mat[2][2]=1;     A.mat[0][1]=2;     Node B;     memset(B.mat,0,sizeof B.mat);     B.mat[0][0]=1;B.mat[1][0]=0;B.mat[2][0]=1;     while(scanf("%lld%lld",&n,&m)!=EOF)     {         mod=m;         Node ans;         memset(ans.mat,0,sizeof ans.mat);         ans=mul(Pow(A,n-1),B);         printf("%lld\n",ans.mat[0][0]);     }     return 0;}