HDU 4651 Partition

qi=((3*i*i)+i)/2

qi=((3*i*i) -i)/2

每个i下面加或减2次

$p(n) = \sum_i (-1)^{i-1} p(n-q_i)$

Partition

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 746 Accepted Submission(s): 438

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

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zhuyuanchen520

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include <map>
 3 #include <queue>
 4 #include <vector>
 5 #include <string>
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 #define maxn 100005
12 #define mod 1000000007
13 #define ll long long
14 #define INF 0x7fffffff
15 int n, m, s;
16 ll f[maxn];
17 void init(){
18     int s = 1;
19     f[0] = 1;
20     for (int i = 1; i <= 100000; i++){
21         ll s = 0;
22         for (int j=1,k = 1,flag=1;j>0; k++,flag=-flag){
23             j = i - (3 * k*k - k) / 2;
24             if (j >= 0)s = (s + (flag*f[j]+mod)%mod) % mod;
25             j = i - (3 * k*k + k) / 2;
26             if (j >= 0)s = (s + (flag*f[j]+mod)%mod) % mod;
27         }
28         f[i] = s;
29     }
30 }
31 int main(){
32     int cas = 1;
33     int t;
34     init();
35     scanf("%d", &t);
36     while (t--){
37         scanf("%d", &n);
38         printf("%d\n", f[n]);
39     }
40     /*while (~scanf("%d", &n)){
41         printf("%I64d\n", f[n]);
42     }*/
43     return 0;
44 }

View Code 2013-10-25 13:44:44

posted @ 2013-10-24 23:56 HaibaraAi 阅读(118) 评论(0) 收藏举报

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