Reading comprehension HDU - 4990

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;

}

InputMulti test cases，each line will contain two integers n and m. Process to end of file.

[Technical Specification]

1<=n, m <= 1000000000OutputFor each case，output an integer，represents the output of above program.Sample Input

1 10
3 100

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 long
using 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)

bcdf
{
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;
}