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HDU 4658 Integer Partition (2013多校6 1004题)

 

Integer Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 169    Accepted Submission(s): 79

Problem Description

 

Given n, k, calculate the number of different (unordered) partitions of n such that no part is repeated k or more times. 

 

 


Input

 

First line, number of test cases, T.  Following are T lines. Each line contains two numbers, n and k. 
1<=n,k,T<=105

 

 


Output

 

T lines, each line contains answer to the responding test case.  Since the numbers can be very large, you should output them modulo 109+7.

 

 


Sample Input

 

4 4 2 4 3 4 4 4 5

 

 


Sample Output

 

2 4 4 5

 

 


Source

 

 

 

 

 

 

欧拉函数的倒数是分割函数母函数,亦即:

p(n)生成函数

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

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)
以上就可以把p(n)求出来了,接下来就来看看本题与之有什么联系呢?
 
首先我们可以写出本题的母函数
 
Σf(n)xˆn=(1+x+x^2+..+x^(k-1))*(1+x^2+x^4+..+x^2(k-1))*..*(1+x^n+..+x^n(k-1));(题目中说 no part is repeated k or more times)
 
     =(1-x^k)*(1-x^2k)*...*(1-x^nk)/((1-x)*(1-x^2)*....*(1-x^n);
 
     =(1-x^k)*(1-x^2k)*...*(1-x^nk)*Σp(n)xˆn;
    
对于(1-x^k)*(1-x^2k)*...*(1-x^nk),可令x^k=y;再利用五边形定理将其打开
 
之后就简单啦!!自己搞一下吧!!
 
 
 1 #include<stdio.h>
 2 typedef __int64 ll;
 3 const int mo=1000000007;
 4 int p[100010];
 5 void pre()
 6 {
 7     p[0]=1;
 8     for(int i=1;i<=100000;i++)
 9     {
10         int t=1,ans=0,kk=1;
11         while(1)
12         {
13             int tmp1,tmp2;
14             tmp1=(3*kk*kk-kk)/2;
15             tmp2=(3*kk*kk+kk)/2;
16             if(tmp1>i)break;
17             ans=(ll)(ans+(ll)t*p[i-tmp1]+mo)%mo;
18             if(tmp2>i)break;
19             ans=(ll)(ans+(ll)t*p[i-tmp2]+mo)%mo;
20             t=-t;
21             kk++;
22         }
23         p[i]=ans;
24     }
25 }
26 ll work(int n,int k)
27 {
28     int i,j;
29     ll ans;
30     ans=p[n];
31     int t=1,kk=1;
32     while(1)
33     {
34         t=-t;
35         ll tmp1,tmp2;
36         tmp1=(ll)k*(3*kk*kk-kk)/2;
37         tmp2=(ll)k*(3*kk*kk+kk)/2;
38         if(tmp1>n)break;
39         ans=(ans+t*p[n-tmp1]+mo)%mo;
40         if(tmp2>n)break;
41         ans=(ans+t*p[n-tmp2]+mo)%mo;
42         kk++;
43     }
44     return ans;
45 }
46 int main()
47 {
48     pre();
49     int T,n,k;
50     scanf("%d",&T);
51     while(T--)
52     {
53         scanf("%d%d",&n,&k);
54         ll ans=work(n,k);
55         printf("%I64d\n",ans);
56     }
57 }
View Code

 

 
posted @ 2013-08-09 20:08  朱群喜_QQ囍_海疯习习  阅读(513)  评论(0编辑  收藏  举报
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