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+7;
int dp[100010];
void init()
{
    memset(dp,0,sizeof(dp));
    dp[0] = 1;
    for(int i = 1;i <= 100000;i++)
    {
        for(int j = 1, r = 1; i - (3 * j * j - j) / 2 >= 0; j++, r *= -1)
        {
            dp[i] += dp[i -(3 * j * j - j) / 2] * r;
            dp[i] %= MOD;
            dp[i] = (dp[i]+MOD)%MOD;
            if( i - (3 * j * j + j) / 2 >= 0 )
            {
                dp[i] += dp[i - (3 * j * j + j) / 2] * r;
                dp[i] %= MOD;
                dp[i] = (dp[i]+MOD)%MOD;
            }

        }



    }
}

int solve(int n,int k)
{
    int ans = dp[n];
    for(int j = 1, r = -1; n - k*(3 * j * j - j) / 2 >= 0; j++, r *= -1)
    {
        ans += dp[n -k*(3 * j * j - j) / 2] * r;
        ans %= MOD;
        ans = (ans+MOD)%MOD;
        if( n - k*(3 * j * j + j) / 2 >= 0 )
        {
            ans += dp[n - k*(3 * j * j + j) / 2] * 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
",solve(n,k));
    }
    return 0;
}