HDU 4651 Partition 整数划分，可重复情况

Partition

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 842    Accepted Submission(s): 478

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

  
    

     
   4 
  
    
  
    

     
   5
  
    
  
    

     
   11 
  
    
  
    

     
   15 
  
    
  
    

     
   19
  
    

Sample Output

  
    

     
   7 
  
    
  
    

     
   56 
  
    
  
    

     
   176 
  
    
  
    

     
   490
  
    

Source

2013 Multi-University Training Contest 5

 

 1  /* *
 2 
 3  1  2   5   7   12   15  22  26
 4  * */
 5 
 6 #include<iostream>
 7 #include<stdio.h>
 8 #include<cstring>
 9 #include<cstdlib>
10  using  namespace std;
11 typedef __int64 LL;
12  const  int maxn = 1e5+ 1;
13 LL  p =  1000000007;
14 LL dp[maxn];
15 LL five1[ 1002];
16 LL five2[ 1002];
17  void init()
18 {
19      int i,j,t;
20      for(LL k= 1;k<= 1000;k++)
21         five1[k] = k*( 3*k- 1)/ 2;
22      for(LL k= 1;k<= 1000;k++)
23         five2[k] = k*( 3*k+ 1)/ 2;
24 
25     dp[ 0]= 1;
26      for(i= 1;i<maxn;i++)
27     {
28         dp[i] =  0;
29         j= 1;
30         t= 1;
31          while( 1)
32         {
33              if(i - five1[t]>= 0)
34             {
35                 dp[i] =dp[i] + j*dp[i-five1[t]] ;
36             }
37              else  break;
38 
39             dp[i] = (dp[i]%p+p)%p;
40 
41              if( i - five2[t]>= 0)
42             {
43                 dp[i] = dp[i] + j*dp[i-five2[t]] ;
44             }
45              else  break;
46 
47             dp[i] = (dp[i]%p+p)%p;
48 
49             j = -j;
50             t++;
51         }
52         dp[i] = (dp[i]%p+p)%p;
53     }
54 }
55  int main()
56 {
57      int T,n;
58     init();
59     scanf( " %d ",&T);
60      while(T--)
61     {
62         scanf( " %d ",&n);
63         printf( " %I64d\n ",dp[n]);
64     }
65      return  0;
66 }