HDU 4602 Partition

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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≤10⁹).

Output

Output the required answer modulo 10⁹+7 for each test case, one per line.

Sample Input

24 25 5

Sample Output

5
1
#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>#include<string>#include<queue>#include<stack>using namespace std;const int maxn=1e9+7;long long T,n,k,ans;long long get(int x){    long long i,j,k;    for (i=2,j=x,k=1;j;j>>=1)    {        if (j&1) (k*=i)%=maxn;        (i*=i)%=maxn;    }    return k;}int main(){    cin>>T;    while (T--)    {        cin>>n>>k;        ans=0;        if (k>=n-1) ans=max(n-k+1,ans);        else         {            ans=n-k+3;            (ans*=get(n-k-2))%=maxn;            }        cout<<ans<<endl;    }    return 0;}