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>
 using namespace std;
 typedef __int64 LL;
 const int maxn = 1e5+;
 LL  p = ;
 LL dp[maxn];
 LL five1[];
 LL five2[];
 void init()
 {
     int i,j,t;
     for(LL k=;k<=;k++)
         five1[k] = k*(*k-)/;
     for(LL k=;k<=;k++)
         five2[k] = k*(*k+)/;
 
     dp[]=;
     for(i=;i<maxn;i++)
     {
         dp[i] = ;
         j=;
         t=;
         while()
         {
             if(i - five1[t]>=)
             {
                 dp[i] =dp[i] + j*dp[i-five1[t]] ;
             }
             else break;
 
             dp[i] = (dp[i]%p+p)%p;
 
             if( i - five2[t]>=)
             {
                 dp[i] = dp[i] + j*dp[i-five2[t]] ;
             }
             else break;
 
             dp[i] = (dp[i]%p+p)%p;
 
             j = -j;
             t++;
         }
         dp[i] = (dp[i]%p+p)%p;
     }
 }
 int main()
 {
     int T,n;
     init();
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         printf("%I64d\n",dp[n]);
     }
     return ;
 }