题意：

告诉你n，m，k。

让你构造k 个数，使得他们的和为n ，最小公倍数是m，求方案数？

思路：

本以为是一个数学，是一个水dp。

我们直接按照他描述的建立状态。

令 dp[i][j][k] , 表示目前选了i 个数，和为j ，最小公倍数是k 的方案数。

那么转移方程就很好写了 dp[i+1][j + a][lcm(a, k)] += dp[i][j][k].

但是得优化一些地方。

1.首先空间上，这样是爆内存的，把第一维改成滚动数组就行了。

第二个，时间上优化。

转移时候你得枚举上一个数是啥，这一个数是啥，这样1000的方了。

我们知道，假如k 个数的最小公倍数是m 的话，那么这k个数都是m 的因子。

2.那这样就好办了。我们先处理m 的因子数量（最多不超过50）.

然后枚举因子进行转移就好了。这就成了50的方了。

3.另外还可以加入一些小的优化。

因为转移到m 只需要因子，那么不如把因子离散化一下。

这样内存就开到了 dp[2][1000][50].

4. 在一点就是 lcm 时候也比较慢，不如预处理出来所有因子的lcm。这样olog 变成了 o1

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;int n, m, k;const int mod = 1000000000 + 7;long long dp[2][1007][50];int gcd(int a,int b){return !b ? a :gcd(b, a%b);
}int lcm(int a,int b){return a*b / gcd(a,b);
}int LCM[1007][1007];int fac[1007];
int cnt = 0;void add(long long& ans, long long x){ans += x;if (ans >= mod) ans -= mod;}
int pos[1007];
int main(){while(~scanf("%d %d %d",&n, &m, &k)){for (int i = 0; i <= n; ++i){for (int j = 0; j < 50; ++j){dp[0][i][j] = 0;}}cnt = 0;for (int i = 1; i <= m; ++i){if (m % i == 0) {fac[cnt++] = i;pos[i] = cnt-1;dp[0][i][pos[i] ] = 1;}}for (int i = 0; i < cnt; ++i){for (int j = 0; j < cnt; ++j){LCM[i][j] = lcm(fac[i], fac[j]);}}for (int i = 1; i < k; ++i){int second = (i & 1);int first = second ^ 1;for (int j = 0; j <= n; ++j){for (int o = 0; o < 50; ++o){dp[second][j][o] = 0LL;}}for (int j = 0; j <= n; ++j){for (int l = 0; l < cnt; ++l){int f = fac[l];if (dp[first][j][pos[f] ]){for (int  m = 0; m < cnt; ++m){int f2 = fac[m];if (j + f2 <= n)add(dp[second][j + f2][pos[LCM[l][m]] ] , dp[first][j][pos[f] ]);}}}}}printf("%I64d\n", dp[!(k & 1) ][n][pos[m] ]);}return 0;
}

Math Magic

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2133 Accepted Submission(s): 709

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

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