HDU 4062 Partition

Partition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1049 Accepted Submission(s): 427

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

 

import java.io.BufferedInputStream;
import java.util.*;

public class Main {
	public static long t1=1000000000+7;
	public static void main(String[] args) {
		Scanner sc = new Scanner(new BufferedInputStream(System.in));
		int t = sc.nextInt();
		for (int i = 0; i < t; i++) {
			int n = sc.nextInt();
			int k = sc.nextInt();
			if (k > n) {
				System.out.println("0");

			} else {
				int b = n - k + 1;
				if (b == 1)
					System.out.println("1");
				else if (b == 2)
					System.out.println("2");
				else if (b == 3)
					System.out.println("5");
				else {
					long num = (power(2,b-1)%t1  + (b - 2) * power(2,b-3))%t1 ;
					num%=t1;
					System.out.println(num);
				}
			}
		}
	}
	 public static long power(int a,int n)
	{
	    if (n==0) return 1;
	    if (n==1) return a;
	     long z=power(a,n/a);
	    if (n%2==0)
	        return z*z%t1;
	    else return z*z*a%t1;
	}  

}