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

A branch B = gcd (a, B) Branch 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 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? RST 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 // today is a long time to understand, tangle, read the code of the great god, only when DP 2/dp [k] [N] [m] is used to indicate the sum of K numbers n, when the minimum public multiple is m, the total number is 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", & N, & M, & K)> 0) 76 solve (); 77 return 0; 78}