HDU 6050 Funny Function —— 2017 Multi-University Training 2

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Funny Function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1213    Accepted Submission(s): 594

Problem Description
Function Fx,ysatisfies:
技术分享

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
 
题目大意:题目给出递推式,对给定的n,m,求F(m,1)
思路:用观察、归纳的方法可以推出:n为奇数时,F(m,1)=(2*(2^n-1)^m+1)/3,
                 n为偶数时,F(m,1)=2*(2^n-1)^m/3.
     用个快速幂加逆元就可以搞定了。难点在于推出公式。
 
AC代码:
 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 const long long MOD=1e9+7;
 6 long long quick_pow(long long a, long long p){
 7     long long ans=1;
 8     while(p){
 9         if(p&1) ans=ans*a%MOD;
10         a=a*a%MOD;
11         p>>=1;
12     }
13     return ans;
14 }
15 int main()
16 {
17     long long n, m;
18     int T;
19     cin>>T;
20     while(T--)
21     {
22         //cout<<‘*‘<<endl;
23         cin>>n>>m;
24         
25         long long res=(quick_pow(2, n)-1)%MOD;
26         res=2*quick_pow(res, m-1)%MOD;
27         if(n&1)
28             res=(res+1)%MOD;
29         res=res*quick_pow(3, MOD-2)%MOD;
30         cout<<res<<endl; 
31     }
32 } 

  个人感觉这种方法不是特别好,这里推荐nicetomeetu的题解, 是用矩阵快速幂做的,公式推的比较详细,可以借鉴一下思路。








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