hdu 4427 Math Magic(DP,4级)

H - Math Magic

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4427

Description

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, ..., A_i, A_i+1,..., A_K) = N
2. LCM (A₁, A₂, ..., A_i, A_i+1,..., A_K) = M
Can you calculate how many kinds of solutions are there for A_i (A_i 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(10⁹ + 7).
You can get more details in the sample and hint below.

 

Sample Input

4 2 23 2 2

 

Sample Output

12

Hint

The first test case: the only solution is (2, 2).The second test case: the solution are (1, 2) and (2, 1). 

 

思路： dp[K][N][M]; K个 ，和，LCM

             dp[K][N][M]=sum(dp[K-1][N-x][y]);LCM(x,y)=M;

     

#include<cstdio>#include<cstring>#include<iostream>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))using namespace std;const int mm=1002;const int mod=1e9+7;int LCM[mm][mm];int dp[2][mm][mm],ppp[mm],pos;int gcd(int a,int b){ int c;  while(b)  {    c=b;b=a%b;a=c;  }  return a;}int lcm(int a,int b){ return a*b/gcd(a,b);}int main(){  FOR(i,1,mm-1)FOR(j,1,mm-1)  LCM[i][j]=lcm(i,j);  int N,M,K;///dp num sum LCM  while(~scanf("%d%d%d",&N,&M,&K))  {    pos=0;    FOR(i,1,M)    if(M%i==0)    ppp[pos++]=i;    int now=0;    FOR(i,0,N)FOR(j,0,pos-1)    dp[now][i][ ppp[j] ]=0;    dp[now][0][1]=1;    FOR(i,1,K)///K次    {      now^=1;      FOR(j,0,N)FOR(k,0,pos-1)      dp[now][j][ ppp[k] ]=0;///clear dp      FOR(j,i-1,N)///至少是1      FOR(k,0,pos-1)      {        if(dp[now^1][j][ ppp[k] ]==0)continue;        FOR(p,0,pos-1)///新加p        {         int x=j+ppp[p];         int y=LCM[ ppp[p] ][ ppp[k] ];         if(x>N||M%y!=0)continue;         dp[now][x][y]+=dp[now^1][j][ ppp[k] ];         if(dp[now][x][y]>=mod)            dp[now][x][y]-=mod;        }      }    }    printf("%d\n",dp[now][N][M]);  }}