2017多校2 1006 Funny Function

Funny Function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1082    Accepted Submission(s): 513

Problem Description

Function Fx,y satisfies:

For given integers N and M,calculate Fm,1 modulo 1e9+7.

 

Input

There is one integer T in the first line.
The next T lines,each line includes two integers N and M .
1<=T<=10000,1<=N,M<2^63.

 

Output

For each given N and M,print the answer in a single line.

 

Sample Input

  
  

   
   2
2 2
3 3
  
  

 

Sample Output

  
  

   
   2
33
  
  

 

这道题可以根据题目给的公式打表，再结合特征根法以及等比等差求和公式来得到最后的公式。 n为奇数 ans=((2*(2^n-1))^(m-1)+1)/3  n为偶数 ans=((2*(2^n-1))^(m-1))/3

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iomanip>
#include<functional>
#include<iostream>
#include<algorithm>
using namespace std;
const long long int MOD=7+1e9;
long long int poww(long long int x,long long int y)
{
    long long int ans=1;
    while(y)
    {
        if(y&1)ans=ans*x%MOD;
        x=x*x%MOD;
        y>>=1;
    }
    return ans%MOD;
}
int main(){
    int t;
    cin>>t;
    while(t--)
    {
        long long int n,m;
        cin>>n>>m;
        long long int  ans;
        if(n%2==1)
         ans=(2*poww(poww(2,n)-1,m-1)%MOD+1)%MOD*poww(3,MOD-2)%MOD;
        else  ans=((2*poww(poww(2,n)-1,m-1))%MOD)*poww(3,MOD-2)%MOD;
        cout<<ans%MOD<<endl;
    }
    return 0;
}