hdu 4427 Math Magic(简单DP注意细节)

Math Magic

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 495    Accepted Submission(s): 210

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). 
    
    
    
    
     
   
   
   
   

 

Source

2012 Asia ChangChun Regional Contest

 

Recommend

zhuyuanchen520

 

题意：

给你k个数的和和k个数的最小公倍数。问你一共有多少满足条件的解。

开始SB了。以为是算组合数。纠结了半天。结果一看样例就凌乱了。

1,2。2,1。是不同的组合也就是说和排序有关。

以后一定要先看完样例 了。。。。

这下就简单了。用三维来表示状态。

dp[i][j][k]。表示长度为i。和为j。最小公倍数为k的方法数。

设a为解的第i+1个数。

那么状态转移就为

dp[i+1][j+a][lcm(a,k)]+=dp[i][j][k]。

lcm为最大公倍数。

由于三维开不下就只能用滚动数组了。

为了节约时间先预处理出1000以内任意两数的最小公倍数。注意一点。一个数学常识。如果这k个数的最小公倍数是

x。那么这k个数肯定都是x的因子。所以先预处理出x的因子。这样就可以有效的减少枚举范围。还有memset要慎用尤其是在这种数量级下。

详细见代码：

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
const int maxn=1010;
const int mod=1000000007;

int dp[2][maxn][maxn];
int lcm[maxn][maxn];//lcm[a][b]为a,b的最大公约数。
int fac[maxn];//存m的因子
int gcd(int a,int b)
{
    int t;
    while(b)
    {
        t=a%b;
        a=b;
        b=t;
    }
    return a;
}
int main()
{
    int n,m,k,cnt,i,j,p,q,e,f,t,v,mul,tt;

    for(i=1; i<maxn; i++)
        for(j=i; j<maxn; j++)//预处理出任意两数的最大公倍数
            lcm[i][j]=lcm[j][i]=i*j/gcd(i,j);
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        cnt=0;
        for(i=1; i<=m; i++)
        {
            if(m%i==0)
                fac[cnt++]=i;
        }
        v=0;
        memset(dp,0,sizeof dp);//记得要初始化。wa了千百回。里面的初始化只初始化了部分
        for(i=0; i<cnt; i++)
            dp[v][fac[i]][fac[i]]=1;
        for(i=1; i<k; i++) //len
        {
            for(e=i; e<=n; e++)//必须手动初始化。不然超时
                for(f=0; f<cnt; f++)
                    dp[v^1][e][fac[f]]=0;
            for(j=i; j<n; j++) //sum
            {
                for(p=0; p<cnt; p++) //mul
                {
                    mul=fac[p];
                    if(!dp[v][j][mul])
                        continue;
                    for(q=0; q<cnt; q++) //a[i]
                    {
                        tt=j+fac[q];
                        if(tt>n)
                            break;
                        t=lcm[mul][fac[q]];
                        dp[v^1][tt][t]+=dp[v][j][mul];
                        dp[v^1][tt][t]%=mod;
                    }
                }
            }
            v^=1;
        }
        printf("%d\n",dp[v][n][m]);
    }
    return 0;
}