zoj3662Math Magic

Math Magic

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Time Limit: 3 Seconds       Memory Limit: 32768 KB

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Yesterday, my teacher taught us about math: +, -, *, /, GCD, LCM... As you know, LCM (Least common multiple) of two positive numbers can be solved easily because of a * b = GCD (a, b) * LCM (a, b).

In class, I raised a new idea: "how to calculate the LCM of K numbers". It's also an easy problem indeed, which only cost me 1 minute to solve it. I raised my hand and told teacher about my outstanding algorithm. Teacher just smiled and smiled...

After class, my teacher gave me a new problem and he wanted me solve it in 1 minute, too. If we know three parameters N, M, K, and two equations:

1. SUM (A₁, A₂, ..., A_i, A_i+1,..., A_K) = N 
2. LCM (A₁, A₂, ..., A_i, A_i+1,..., A_K) = M

Can you calculate how many kinds of solutions are there for A_i (A_i are all positive numbers). I began to roll cold sweat but teacher just smiled and smiled.

Can you solve this problem in 1 minute?

Input

There are multiple test cases.

Each test case contains three integers N, M, K. (1 ≤ N, M ≤ 1,000, 1 ≤ K ≤ 100)

Output

For each test case, output an integer indicating the number of solution modulo 1,000,000,007(1e9 + 7).

You can get more details in the sample and hint below.

Sample Input

4 2 2
3 2 2

Sample Output

1
2

Hint

The first test case: the only solution is (2, 2).

The second test case: the solution are (1, 2) and (2, 1).

这题时间卡的真紧啊！

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
#define MAXN 1005
#define mod 1000000007
int lca[MAXN][MAXN],dp[2][MAXN][MAXN],vec[MAXN];
int gcd(int a,int b)
{
    if(a==0)return b;
    return gcd(b%a,a);
}
int main()
{
    int n,m,k,i,j,now,no,k1,j1,ans,ii;
    for(i=1;i<=1000;i++)
        for(j=i;j<=1000;j++)
            lca[j][i]=lca[i][j]=i/gcd(i,j)*j;
    while(scanf("%d%d%d",&n,&m,&no)!=EOF)
    {
        now=0;
        //memset(dp,0,sizeof(dp));
        ans=0;
        vec[ans++]=1;
        for(i=2;i<=m;i++)
        {
            if(m%i==0)
            vec[ans++]=i;
        }
        for(ii=0;ii<=n;ii++)
            for(j=0;j<ans;j++)
                dp[now][ii][vec[j]]=0;
            dp[now][0][1]=1;
        for(i=0;i<=no-1;i++)
        {
            now=now^1;
            for(ii=0;ii<=n;ii++)
                for(j=0;j<ans;j++)
                dp[now][ii][vec[j]]=0;
            for(j=i;j<=n;j++)
                for(int j2=0;j2<ans;j2++)
                {
                    k=vec[j2];
                    if(dp[now^1][j][k]==0)
                    continue;
                      for(int jj1=0;jj1<ans;jj1++)
                    {

                        j1=vec[jj1];
                        if(j1+j>n)
                        break;
                        k1=lca[k][j1];
                       if(k1>m||m%k1!=0)
                        continue;
                        dp[now][j1+j][k1]+=dp[now^1][j][k];

                        dp[now][j1+j][k1]%=mod;
                    }
                }

        }
        printf("%d\n",dp[now][n][m]%mod);
    }
    return 0;
}