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≤10⁹).

Output

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

Sample Input

24 25 5

Sample Output

51

 思路：

把1~4的每个元素的个数列出来后就可以简单看出其中的固定规律

   1 2 3  4

 11 2 5 12

2  1 2 5

3    1 2

4       1

 

所以数列符合a_1 = 1, 2, 5, 12 。。。。。a_n = 2*f(n-1)+2^(n-3)

 

最后问题推导出的公式：a_n = 2^(n-1) + (n-2)*2^(n-3);

 

 

 

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;  }  }