2017 Multi-University Training Contest 2 && HDOJ 6050 Funny Function 【思维+快速幂】

Funny Function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit:32768/32768 K (Java/Others)

Total Submission(s): 284    Accepted Submission(s): 121


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

【题意】给出递推公式，递推公式与n有关，求F[m][1]的值。

【思路】这道题我们可以经过打表发现最终的递推公式与n有关：

当n为偶数时：F(m,1)=(2^n -1)^(m-1)*2/3
当n为奇数时：F(m,1)=((2^n -1)^(m-1)*2+1)/3

这样的话可以直接利用快速幂求解，值得注意的是，这里涉及到除法取膜，由于除法取膜不满足公式，所以要求3对于1e9+7的逆元，由拓展欧几里得算法或者费马小定理可以得到逆元为333333336，代入即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int T;scanf("%d",&T);while(T--)

typedef unsigned long long ll;
const int maxn = 100005;
const ll mod = 1e9+7;
const ll inv = 333333336;
const int INF = 0x3f3f3f;
const double eps = 1e-9;

ll fast_mod(ll a,ll b,ll Mod)
{
ll ans=1;
a%=Mod;
while(b)
{
if(b&1)
ans=(ans*a)%Mod;
b>>=1;
a=(a*a)%Mod;
}
return ans;
}

int main()
{
ll n,m;
rush()
{
scanf("%I64d%I64d",&n,&m);
if(m==1)
{
puts("1");
continue;
}
ll temp=(fast_mod(2,n,mod)-1+mod)%mod;
ll ans;
if(n&1)
{
ans=(2*fast_mod(temp,m-1,mod)+1)*inv%mod;
}
else
{
ans=2*fast_mod(temp,m-1,mod)*inv%mod;
}
printf("%I64d\n",ans);
}
return 0;
}