UVALive 6073 Math Magic

                                              6073 Math Magic
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, . . . , Ai, Ai+1, . . . , AK) = N
          2. LCM(A1, A2, . . . , Ai, Ai+1, . . . , AK) = M

Can you calculate how many kinds of solutions are there for Ai (Ai 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.
Hint:
  The first test case: the only solution is (2, 2).
  The second test case: the solution are (1, 2) and (2, 1).

Sample Input
4 2 2
3 2 2

Sample Output
1
2

//今天算是长见识了,纠结,看了大神的代码,才知道用dp
//dp[k][n][m]表示由k个数组成的和为n,最小公倍数为m的情况总数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1005;
const int mod = 1000000007;
int n, m, k;
int lcm[maxn][maxn];
int dp[2][maxn][maxn];
int fact[maxn], cnt;

int GCD(int a, int b)
{
    return b==0?a:GCD(b, a%b);
}

int LCM(int a, int b)
{
    return a / GCD(a,b) * b;
}

void init()
{
    for(int i = 1; i <=1000; i++)
        for(int j = 1; j<=i; j++)
            lcm[j][i] = lcm[i][j] = LCM(i, j);
}

void solve()
{
    cnt = 0;
    for(int i = 1; i<=m; i++)
        if(m%i==0) fact[cnt++] = i;

    int now = 0;
    memset(dp[now], 0, sizeof(dp[now]));
    for(int i = 0; i<cnt; i++)
        dp[now][fact[i]][fact[i]] = 1;

    for(int i = 1; i<k; i++)
    {
        now ^= 1;
        for(int p=1; p<=n; p++)
            for(int q=0; q<cnt; q++)
            {
                dp[now][p][fact[q]] = 0;
            }

        for(int p=1; p<=n; p++)
        {
            for(int q=0; q<cnt; q++)
            {
                if(dp[now^1][p][fact[q]]==0) continue;
                for(int j=0; j<cnt; j++)
                {
                    int now_sum = p + fact[j];
                    if(now_sum>n) continue;
                    int now_lcm = lcm[fact[q]][fact[j]];
                    dp[now][now_sum][now_lcm] += dp[now^1][p][fact[q]];//
                    dp[now][now_sum][now_lcm] %= mod;//
                }
            }
        }
    }
    printf("%d
",dp[now][n][m]);
}

int main()
{
    init();
    while(scanf("%d%d%d", &n, &m, &k)>0)
        solve();
    return 0;
}