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,ysatisfies:

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

22 23 3

 

Sample Output

233

 

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;}