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): 22    Accepted Submission(s): 15

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

2013 Multi-University Training Contest 6

 

Recommend

zhuyuanchen520

 

 

 

跟上次多校求数的划分很类似。

 

 

所谓的五边形数定理还没有搞懂。

 

 

先贴个代码先，胡搞弄过去的

 

 

 /*
  * Author:  kuangbin
  * Created Time:  2013/8/8 11:53:35
  * File Name: 1004.cpp
  */
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <time.h>
 using namespace std;
 const int MOD = 1e9+;
 int dp[];
 void init()
 {
     memset(dp,,sizeof(dp));
     dp[] = ;
     for(int i = ;i <= ;i++)
     {
         for(int j = , r = ; i - ( * j * j - j) /  >= ; j++, r *= -)
         {
             dp[i] += dp[i -( * j * j - j) / ] * r;
             dp[i] %= MOD;
             dp[i] = (dp[i]+MOD)%MOD;
             if( i - ( * j * j + j) /  >=  )
             {
                 dp[i] += dp[i - ( * j * j + j) / ] * r;
                 dp[i] %= MOD;
                 dp[i] = (dp[i]+MOD)%MOD;
             }

         }

     }
 }

 int solve(int n,int k)
 {
     int ans = dp[n];
     for(int j = , r = -; n - k*( * j * j - j) /  >= ; j++, r *= -)
     {
         ans += dp[n -k*( * j * j - j) / ] * r;
         ans %= MOD;
         ans = (ans+MOD)%MOD;
         if( n - k*( * j * j + j) /  >=  )
         {
             ans += dp[n - k*( * j * j + j) / ] * r;
             ans %= MOD;
             ans = (ans+MOD)%MOD;
         }

     }
     return ans;
 }

 int main()
 {
     init();
     int T;
     int n,k;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&k);
         printf("%d\n",solve(n,k));
     }
     return ;
 }