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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 ?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 //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 }
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原文地址：http://www.cnblogs.com/zyx1314/p/3830708.html
