连接

hdu 4651 Partition(整数拆分+五边形数)

 

Partition

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 462    Accepted Submission(s): 262

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 ≤ 105) you need to consider.

 

 


Output

 

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

 

 


Sample Input

 

4 5 11 15 19

 

 


Sample Output

 

7 56 176 490

 

 


Source

 

2013 Multi-University Training Contest 5

 

 


Recommend

 

zhuyuanchen520

 

 

 

 

 

 

 

欧拉函数的倒数是分割函数的母函数，亦即：

p(n)的生成函数是

\sum_{n=0}^\infty p(n)x^n = \prod_{k=1}^\infty \left(\frac {1}{1-x^k} \right)   （1）
再

利用五边形数定理可得到以下的展开式：

\prod_{k=1}^\infty (1-x^k)=\sum_{i=-\infty}^\infty(-1)^ix^{i(3i-1)/2}. （2）
将（2）式带入（1）式，并乘到（1）式的左边，进行展开，合并同类项，根据非常数项的系数为0！！
 
即将p(n)生成函数配合五边形数定理，可以得到以下的递归关系式
p(n) = \sum_i (-1)^{i-1} p(n-q_i)

 

 

 

 1 #include<stdio.h>
 2 typedef long long ll;
 3 const int mo=1000000007;
 4 ll p[100010];
 5 void pre()//打表，欧拉函数的倒数是分割函数的母函数！！！ 
 6 {
 7     p[0]=1;
 8     for(ll i=1;i<=100000;i++)
 9     {
10         ll t=1,ans=0,kk=1;
11         while(1)
12         {
13             ll tmp1,tmp2;
14             tmp1=(3*kk*kk-kk)/2;
15             tmp2=(3*kk*kk+kk)/2;
16             if(tmp1>i)break;
17             ans=(ans+t*p[i-tmp1]+mo)%mo;
18             if(tmp2>i)break;
19             ans=(ans+t*p[i-tmp2]+mo)%mo;
20             t=-t;
21             kk++;
22         }
23         p[i]=ans;
24     }
25 }
26 int main()
27 {
28     pre();
29     int T,n;
30     scanf("%d",&T);
31     while(T--)
32     {
33         scanf("%d",&n);
34         printf("%lld\n",p[n]);
35     }
36 }
View Code

 

 

 

posted @ 2013-08-08 08:52  朱群喜_QQ囍_海疯习习  阅读(354)  评论(0)  编辑  收藏  举报
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