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 ≤ 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

 

 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 }