HDU 6050 17多校2 Funny Function（数学+乘法逆元）

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

2

2 2

3 3

 

Sample Output

2

33

数学题，反正我不会。。。数学基础太差了，直接用大佬的数学公式写的

参考博客：http://www.cnblogs.com/Just--Do--It/p/7248089.html

主要是学到了取余和除的话需要求逆元！！可以用费马小定理求，目前我只会这个

继续加油！

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string.h>
#include<cmath>
#include<math.h>
using namespace std;

#define MOD 1000000000+7

long long quick_mod(long long a,long long b)//快速幂，复杂度log2n
{
    long long ans=1;
    while(b)
    {
        if(b&1)
        {
            ans*=a;
            ans%=MOD;
            b--;
        }
        b/=2;
        a*=a;
        a%=MOD;
    }
    return ans;
}

long long inv(long long x)
{
    return quick_mod(x,MOD-2);//根据费马小定理求逆元，3*（3^(mod-2)）=1
}

int main()
{
    int T;
    scanf("%d",&T);
    long long n,m,ans;
    while(T--)
    {
        scanf("%lld%lld",&n,&m);
        if(n%2==0)
            ans=(quick_mod(quick_mod(2,n)-1,m-1)*2)*inv(3);
        else
            ans=(quick_mod(quick_mod(2,n)-1,m-1)*2+1)*inv(3);
        ans%=MOD;
        printf("%lld
",ans);
    }
    return true;
}