ZOJ 3662 Math Magic

最新推荐文章于 2020-02-07 11:05:28 发布

baymax520

最新推荐文章于 2020-02-07 11:05:28 发布

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分类专栏：背包dp 文章标签：背包dp

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本文探讨了一道数学与编程结合的问题，即给定数字N、M和K，求解K个正整数之和为N且最小公倍数为M的方案数。通过筛选M的所有因子并使用滚动数组的01背包算法，文章提供了一个高效解决方案。

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Math Magic

Time Limit: 3 Seconds Memory Limit: 32768 KB

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.

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).

Contest: The 2012 ACM-ICPC Asia Changchun Regional Contest

题意：给定数字N和M，K，求K个数字和为N并且最小公倍数为M的方案数。

思路：最小公倍数为M的话那些数字肯定就是M的因子了，我们只需要把M的因子都挑出来，然后将其作为物品进行背包即可，但是有个问题，不做处理直接存的话是会爆内存的，想一想背包九讲当中01背包二维降为一位的想法，挑选的前i个数是由前i-1个数转移来的，那么就可以用滚动数组来进行背包了。dp[i][j][k]表示前i个数字和为j最小公倍数为k的方案数。

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
const int mod=1e9+7;
int dp[2][1005][1005],LCM[1005][1005],num[105];
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int lcm(int a,int b)
{
    return a/gcd(a,b)*b;
}
int main()
{
    int n,m,k;
    for(int i=1;i<=1000;i++)
        for(int j=1;j<=1000;j++)
            LCM[i][j]=lcm(i,j);
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        int cnt=0;
        for(int i=1;i<=m;i++)
        {
            if(m%i==0)
                num[cnt++]=i;
        }
        int now=0;
        memset(dp,0,sizeof(dp));
        dp[now][0][1]=1;
        for(int t=1;t<=k;t++)
        {
            now=now^1;
            for(int i=0;i<=n;i++)
                for(int j=0;j<cnt;j++)
                    dp[now][i][num[j]]=0;
            for(int i=0;i<=n;i++)
            {
                for(int 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;
}