Math Magic（完全背包）

Math Magic

Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

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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(1e9 + 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).

 

 

题意：

给出n,m,k,问k个数的和为n，最小公倍数为m的情况有几种

思路：

因为最小公倍数为m，可以知道这些数必然是m的因子，那么我们只需要选出这所有的因子，拿这些因子来背包就可以了

dp[now][i][j]表示当前状态下，和为i，最小公倍数为j的解的个数。递推K次就出答案了。

注意需要优化！！！

详见代码

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
#define mod 1000000007

int num[1000];
int dp[2][1010][1010];
int LCM[1010][1010];

int gcd(int a,int b)//最大公约数
{
    if(b==0) return a;
    return gcd(b,a%b);
}

int lcm(int a,int b)//最小公倍数
{
    return (a*b/gcd(a,b));
}


int main()
{
    int n,m,k;
    int i,j;
    for(i=1;i<=1000;i++)//预处理，前1000的最小公倍数
    {
        for(j=1;j<=1000;j++)
        {
            LCM[i][j]=lcm(i,j);
        }
    }
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        int cnt=0;
        //因为最小公倍数m已知，所以Ai必定是他的因子
        for(i=1;i<=m;i++)
        {
            if(m%i==0)
                num[cnt++]=i;
        }

        //dp[now][i][j]now表示当前状态下，和为i,最小公倍数为j的解的个数。递推K次就出答案了。
        int now=0;
        //memset(dp[nom],0,sizeof(dp[nom]));
        for(i=0;i<=n;i++)
        {
            for(j=0;j<cnt;j++)
            {
                //初始化，和为i，最小公倍数是num[j]的
                dp[now][i][num[j]]=0;
            }
        }
        dp[0][0][1]=1;

        for(int t=1;t<=k;t++)
        {
            now^=1;
            for(i=0;i<=n;i++)
            {
                for(j=0;j<cnt;j++)
                {
                    dp[now][i][num[j]]=0;
                }
            }

            for(i=t-1;i<=n;i++)
            {
                for(j=0;j<cnt;j++)
                {
                    if(dp[now^1][i][num[j]]==0)continue;
                    for(int p=0;p<cnt;p++)
                    {
                        int x=i+num[p];
                        int y=LCM[num[j]][num[p]];
                        if(x>n||m%y!=0) continue;
                        dp[now][x][y]+=dp[now^1][i][num[j]];
                        dp[now][x][y]%=mod;
                    }
                }
            }
        }
        printf("%d\n",dp[now][n][m]);
    }
    return 0;
}