hdu4602(快速幂)

Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 258    Accepted Submission(s): 127

Problem Description

Define f(n) as the number of ways to perform n in format of the sum of some positive integers. For instance, when n=4, we have
  4=1+1+1+1
  4=1+1+2
  4=1+2+1
  4=2+1+1
  4=1+3
  4=2+2
  4=3+1
  4=4
totally 8 ways. Actually, we will have f(n)=2 ^(n-1) after observations.
Given a pair of integers n and k, your task is to figure out how many times that the integer k occurs in such 2 ^(n-1) ways. In the example above, number 1 occurs for 12 times, while number 4 only occurs once.

 

Input

The first line contains a single integer T(1≤T≤10000), indicating the number of test cases.
Each test case contains two integers n and k(1≤n,k≤10 ⁹).

 

Output

Output the required answer modulo 10 ⁹+7 for each test case, one per line.

 

Sample Input

   
   
   
   

    
    
    
    2
4 2
5 5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    5
1
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    本题在比赛的时候出答案好快，但一时又想不出缘由，于是找规律
   
   
   
   
   
   
   
   

    
    
    
    1,2,5,12,28,64·······
   
   
   
   
   
   
   
   

    
    
    
    1,2,2*2+2^0,2*5+2^1,2*12+2^2······
   
   
   
   
   
   
   
   

    
    
    
    于是想到递推式：
   
   
   
   
   
   
   
   

    
    
    
       a[0]=1,a[1]=2
   
   
   
   
   
   
   
   

    
    
    
       a[i]=2*a[i-1]+2^(i-2),i>1，i=n-k
   
   
   
   
   
   
   
   

    
    
    
    由于1≤n,k≤10
    
    
    
    9很大，直到递推式还不能解决问题，需要得到通式
   
   
   
   
   
   
   
   

    
    
    
       a[0]=1,a[1]=2
    
    
    
    

      a[n-k]=(n-k+3)*2^(n-k-2),n-k>1 
    
    
    
    
    

      由于n-k-2可能很大，需要用快速幂计算 
    
    
    
    
    
 
     
#include<cstdio>
#include<iostream>
using namespace std;
const __int64 mod=1000000000+7;


__int64 optimized_pow_n(__int64 x, __int64 n)
//快速幂x^n
{
    __int64  pw = 1;
    while (n > 0)
	{
        if (n&1)       
            pw *= x;
		 pw=pw%mod;
        x*=x;
		 x=x%mod;
        n>>= 1;     
    }
    return pw%mod;
}

int main()
{
	int t,n,k;
	cin>>t;
	while(t--)
	{
		scanf("%d%d",&n,&k);
		if(k>n)
			printf("0\n");
		else if(n==k)
			printf("1\n");
		else if(n-1==k)
			printf("2\n");
		else 
		{
			printf("%I64d\n",(3+n-k)*optimized_pow_n(2,(__int64)(n-k-2))%mod);
		}
	}     
	return 0;
}