hdu 4602 Partition （概率方法）

Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2472    Accepted Submission(s): 978

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
   
     
   
     


   
     

对于1 <= k < n，我们能够等效为n个点排成一列，并取出当中的连续k个，这连续的看K个两端断开；

1、若取得是这K个点包含端点（我们仅仅考虑一个端点的情况），还剩下（n-k-1）个间隔，

每一个间隔有断开和闭合两种状态，故有2^（n-k-1），最后乘以2；

2、若取得是这K个点不包含端点，这连续的K个点有（n-k-1）种取法，还剩下（n-k-2）个间隔，

故有2^（n-k-2）*（n-k-1）；

总计2 ∗ 2^(n – k − 1) + 2^(n – k − 2) ∗ (n – k − 1) = (n – k + 3) * 2^(n – k − 2)。 

#include"stdio.h"
#include"string.h"
#include"queue"
#include"vector"
#include"algorithm"
using namespace std;
#define LL __int64
const int mod=1000000007;
int main()
{
    int T,i,n,k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        if(n<k)
            printf("0\n");
        else if(n==k)
            printf("1\n");
        else
        {
            LL d=n-k,s1,t;
            if(d==1) printf("2\n");
            else
            {
                s1=d+3;
                d-=2;
                LL aa=2,tmp=1;
                while(d)
                {
                    if(d&1)
                        tmp*=aa;
                    d/=2;
                    aa=(aa*aa)%mod;
                    tmp%=mod;
                }
                printf("%I64d\n",(tmp*s1)%mod);
            }
        }

    }
    return 0;
}