HDU 4602 Partition (整数拆分&找规律&快速幂取模)

Partition

http://acm.hdu.edu.cn/showproblem.php?pid=4602

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)



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

2
4 2
5 5

Sample Output

5
1

1. 观察发现n的分拆中n出现了1次，n-1出现了2次，n-2出现了5次，……，这一发现是否正确？

这样想：n出现了1次不用说；n-1出现了2次是因为剩下那个数是1，它只能和n-1有两种组合，或者当n=2时变成1+1，那也有2个数；n-2出现了5次是因为剩下的数2可以与n-2有3+2种组合，……

那么，记a1=1,a2=2,a3=5,...，我们的目标就是求出an。

2. 如何求出an？

利用公式：an=Sn-Sn-1

3. 如何求出Sn？

考虑将n分成k个正数，由插空法知有C(n-1,k-1)种分法（或者：这等价于方程x1+x2+...+xk = n-k的非负解的个数，亦等价于从k种不同的物品中允许重复地选出n-k个物品的方法数），于是求出总个数Sn：



4. 注意n,k可能较大，要使用快速幂取模。

完整代码：

/*31ms,228KB*/

#include<cstdio>
const int mod = 1000000007;

__int64 pow(__int64 a, __int64 b)///a^b % mod
{
	__int64 r = 1, base = a;
	while (b)
	{
		if (b & 1)
			r = r * base % mod;
		base = base * base % mod;
		b >>= 1;
	}
	return r;
}

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