hdu 4658 Integer Partition(整数拆分变形)

Integer Partition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 503 Accepted Submission(s): 230



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

题目分析：

这道题很容易看出是整数拆分，也就是将母函数利用五边形定理优化

推倒过程直接复制别人的了。。。。见谅

Hdu 4658 要求拆分的数中每个数出现的次数不能大于等于k次，则



代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define MAX 100007

using namespace std;

int n,k,t;

const int MOD = 1e9+7;

int dp[MAX];

int five ( int x )
{
    return (3*x*x-x)/2;
}

void init ( )
{
    memset ( dp , 0 , sizeof ( dp ) );
    dp[0] = 1;
    for ( int i = 1 ; i < MAX ; i++ )
    {
        for ( int j = 1 ; ; j++ )
        {
            int k = five(j);
            if ( k > i ) break;
            if ( j&1 )
                dp[i] += dp[i-k];
            else dp[i] -= dp[i-k];
            dp[i] = dp[i]%MOD;
            k = five(-1*j);
            if ( k > i ) break;
            if ( j&1 )
                dp[i] += dp[i-k];
            else dp[i] -= dp[i-k];
            dp[i] = dp[i]%MOD;
        }
        dp[i] = (dp[i]%MOD+MOD)%MOD;
    }
}

int solve ( int n , int k )
{
    int ans = dp
;
    for ( int i = 1; ; i++ )
    {
       int j = five(i)*k;
       if ( j > n ) break;
       if ( i&1 )
           ans -= dp[n-j];
       else ans += dp[n-j];
       ans%=MOD;
       j = five(-1*i)*k;
       if ( j > n ) break;
       if ( i&1 )
           ans -= dp[n-j];
       else ans += dp[n-j];
       ans %=MOD;
    }
    return (ans%MOD+MOD)%MOD;
}

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