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

 

 1 //今天算是长见识了,纠结,看了大神的代码,才知道用dp

 2 //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

 3 

 4 #include <iostream>

 5 #include <cstdio>

 6 #include <cstring>

 7 #include <algorithm>

 8 using namespace std;

 9 const int maxn = 1005;

10 const int mod = 1000000007;

11 int n, m, k;

12 int lcm[maxn][maxn];

13 int dp[2][maxn][maxn];

14 int fact[maxn], cnt;

15 

16 int GCD(int a, int b)

17 {

18     return b==0?a:GCD(b, a%b);

19 }

20 

21 int LCM(int a, int b)

22 {

23     return a / GCD(a,b) * b;

24 }

25 

26 void init()

27 {

28     for(int i = 1; i <=1000; i++)

29         for(int j = 1; j<=i; j++)

30             lcm[j][i] = lcm[i][j] = LCM(i, j);

31 }

32 

33 void solve()

34 {

35     cnt = 0;

36     for(int i = 1; i<=m; i++)

37         if(m%i==0) fact[cnt++] = i;

38 

39     int now = 0;

40     memset(dp[now], 0, sizeof(dp[now]));

41     for(int i = 0; i<cnt; i++)

42         dp[now][fact[i]][fact[i]] = 1;

43 

44     for(int i = 1; i<k; i++)

45     {

46         now ^= 1;

47         for(int p=1; p<=n; p++)

48             for(int q=0; q<cnt; q++)

49             {

50                 dp[now][p][fact[q]] = 0;

51             }

52 

53         for(int p=1; p<=n; p++)

54         {

55             for(int q=0; q<cnt; q++)

56             {

57                 if(dp[now^1][p][fact[q]]==0) continue;

58                 for(int j=0; j<cnt; j++)

59                 {

60                     int now_sum = p + fact[j];

61                     if(now_sum>n) continue;

62                     int now_lcm = lcm[fact[q]][fact[j]];

63                     dp[now][now_sum][now_lcm] += dp[now^1][p][fact[q]];//

64                     dp[now][now_sum][now_lcm] %= mod;//

65                 }

66             }

67         }

68     }

69     printf("%d\n",dp[now][n][m]);

70 }

71 

72 int main()

73 {

74     init();

75     while(scanf("%d%d%d", &n, &m, &k)>0)

76         solve();

77     return 0;

78 }