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 = ;
 const int mod = ;
 int n, m, k;
 int lcm[maxn][maxn];
 int dp[][maxn][maxn];
 int fact[maxn], cnt;

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

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

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

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

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

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

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

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