HDU 4651 2013多校联合第5场 Partition 数论

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本文探讨了一种利用五边形数定理解决整数分拆问题的高效算法，并提供了具体的实现代码，旨在帮助读者理解如何计算给定整数n的所有可能分拆方式的数量。

Partition

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 422 Accepted Submission(s): 242

Problem Description

How many ways can the numbers 1 to 15 be added together to make 15? The technical term for what you are asking is the "number of partition" which is often called P(n). A partition of n is a collection of positive integers (not necessarily distinct) whose sum equals n.

Now, I will give you a number n, and please tell me P(n) mod 1000000007.

Input

The first line contains a number T(1 ≤ T ≤ 100), which is the number of the case number. The next T lines, each line contains a number n(1 ≤ n ≤ 10 ⁵) you need to consider.

Output

For each n, output P(n) in a single line.

Sample Input

Sample Output

題意：整数分拆

思路：此题坑了，虽然很快想出了dp 的解法确发现根本开不了如此之大的数组。方法是：dp[n][k]代表k个数加起来等于n的种数。转移方程很容易dp[n][k]=dp[n-1][k-1]+dp[n-k][k]。本题正解用的是五边形数定理：（摘wiki百科里面的一段）

$(1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} + \cdots)(1 + p(1)x + p(2)x^2 + p(3)x^3 + \cdots)=1$

考虑 $x^n$ 项的系数，在 n>0 时，等式右侧的系数均为0，比较等式二侧的系数，可得

$p(n) - p(n-1) - p(n-2) + p(n-5) + p(n-7) + \cdots=0$

因此可得到分割函数p(n)的递归式

$p(n) = p(n-1) + p(n-2) - p(n-5) - p(n-7) + \cdots$

其中规律是两项两项的加减变化。然后里面是n-pi，其中pi是 $p_n = \frac{3n^2 \pm n}{2}$ .其他细节可以看代码。

/*
 * File   :tmp.cpp
 * Author :Kevin Tan
 * Source :ZJNU
 *
 * 2013年8月7日,下午7:16:41
 */

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<climits>
#include<utility>
#include<cctype>
#include<iomanip>
#include<string>
#include<map>
#include<deque>
#include<queue>
#include<set>
#include<vector>
#include<iterator>
using namespace std;
#define MAX 100005
#define MOD 1000000007
int q[MAX], p[MAX];
int main(int argc, char **argv) {
	q[0] = 0;
	int k = 1;
	for (int i = 1; q[k - 1] <= MAX; i++) {
		q[k++] = (3 * i * i - i) / 2;
		q[k++] = (3 * i * i + i) / 2;
	}
	p[0] = 1;
	for (int i = 1; i <= MAX; i++) {
		p[i] = 0;
		for (int j = 1; q[j] <= i; j++) {
			if (((j - 1) >> 1) & 1) p[i] = (p[i] - p[i - q[j]]) % MOD;//(j-1)>>1 &1是符号两位两位的变
			else p[i] = (p[i] + p[i - q[j]]) % MOD;
			if (p[i] < 0) p[i] += MOD;
		}
	}
	int T;
	scanf("%d", &T);
	while (T--) {
		int n;
		scanf("%d", &n);
		printf("%d\n", p[n]);
	}
	return 0;
}