HDU 4427 Math Magic【dp+优化+滚动数组】【好题】

Math Magic

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2084    Accepted Submission(s): 692

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 (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 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

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。

因为开不了那么大的数组，因此要用滚动数组。

为了节约时间先预处理出1000以内任意两数的最小公倍数。

同时在循环的时候要注意剪枝（if (dp[(j-1)&1][i][last] == 0) continue;//如果上一状态为0则不用计算上一步）

#include<iostream>	
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<bitset>
#include<numeric>
#include<vector>
#include<string>
#include<iterator>
#include<cstring>
#include<functional>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;

const int maxn = 10010;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);

typedef pair<int, int> P;
typedef long long ll;
typedef unsigned long long ull;

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 N, M, n;
int dp[2][1000 + 5][1000 + 5];
int LCM[1000 + 5][1000 + 5];
int fac[1000 + 5];

int main()
{
	for (int i = 1; i <= 1000; i++)
	{
		for (int j = i; j <= 1000; j++)
		{
			LCM[i][j] = LCM[j][i] = lcm(i, j);
		}
	}
 	while (scanf("%d %d %d", &N, &M, &n) != EOF)
	{
		int cnt = 0;
		for (int i = 1; i <= M; i++)
		{
			if (M%i == 0) fac[cnt++] = i;
		}

		ms(dp, 0);

		for (int i = 0; i < cnt; i++)
		{
			dp[0][fac[i]][fac[i]] = 1;
		}

		for (int j = 1; j < n; j++)  //当前选的数  从j推j+1 故只需要推到n-1
		{
			ms(dp[j & 1], 0);
			for (int i = j; i <= N; i++)//枚举和 j最小为1所以和最小只可能为1
			{
				for (int k = 0; k < cnt; k++)//枚举上一状态公倍数
				{
					int last = fac[k];
					if (dp[(j-1)&1][i][last] == 0)	continue;//如果上一状态为0则不用计算上一步
					for (int p = 0; p < cnt; p++)//枚举公因子	
					{
						int cur = fac[p];
						if (i+cur<=N)//保证i+cur在范围内
							dp[j & 1][i + cur][LCM[last][cur]] = (dp[j & 1][i + cur][LCM[last][cur]] + dp[(j-1)&1][i][last]) % 1000000007;
					}
				}
			}
		}
		printf("%d
", dp[(n-1)& 1][N][M]);
	}
}