2012年长春站H题 ||hdu 4427 dp

http://acm.hdu.edu.cn/showproblem.php?pid=4427

Problem 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 (A ₁, A ₂, ..., A _i, A _i+1,..., A _K) = N
2. LCM (A ₁, A ₂, ..., 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(10 ⁹ + 7).
You can get more details in the sample and hint below.

 

Sample Input

   
   
   
   

    
    
    
    4 2 2
3 2 2
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
2


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

解题思路：dp[i][j][k]表示长度为i和为j且最小公倍数为k的序列有多少种。防止超时用到了很多优化，详见代码注释。

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;

const int MOD=1e9+7;
const int N=1050;
int dp[2][N][N],LCM[N][N],num[N];//dp[i][j][k]:长度为i，和为j，最小公倍数为k的种类有多少
int n,m,k;

int gcd(int a,int b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}

int lcm(int a,int b)
{
    return a/gcd(a,b)*b;
}

int main()
{
    for(int i=1;i<=1000;i++)//求1000内任意两个数的最小公倍数
            for(int j=1;j<=1000;j++)
                 LCM[i][j]=lcm(i,j);
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        int cnt=0;
        for(int i=1;i<=m;i++)//每一个Ai一定是m的因子，求出m的所有因子
            if(m%i==0)
                num[cnt++]=i;
        int now=0;
        for(int i=1;i<=n;i++)//滚动数组清零
            for(int j=0;j<cnt;j++)
                dp[now][i][num[j]]=0;
        dp[now][0][1]=1;
        for(int t=1;t<=k;t++)
        {
            now^=1;
            for(int i=1;i<=n;i++)// 滚动数组清零
               for(int j=0;j<cnt;j++)
                    dp[now][i][num[j]]=0;
            for(int i=t-1;i<=n;i++)
                for(int j=0;j<cnt;j++)
                {
                     if(dp[now^1][i][num[j]]==0)//优化
                           continue;
                     for(int p=0;p<cnt;p++)
                     {
                         int x=i+num[p];
                         int y=LCM[num[j]][num[p]];
                         if(y>m||x>n)continue;//最小公倍数大于m，长度大于n，根本不用再往下计算
                         dp[now][x][y]+=dp[now^1][i][num[j]];
                         dp[now][x][y]%=MOD;
                     }
                }
        }
        printf("%d\n",dp[now][n][m]);
    }
    return 0;
}