HOJ 1096 Divided Product (DFS）

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

Given two positive integers N and M, please divide N into several integers A1, A2, ..., Ak (k >= 1), so that:

1. 0 < A1 < A2 < ... < Ak;

2. A1 + A2 + ... + Ak = N;

3. A1, A2, ..., Ak are different with each other;

4. The product of them P = A1 * A2 * ... * Ak is a multiple of M;

How many different ways can you achieve this goal?

输入

Two integers N and M. 1 <= N <= 100, 1 <= M <= 50.

输出

Output one integer -- the number of different ways to achieve this goal, module 1,000,000,007.

样例提示

There are 4 different ways to achieve this goal for the sample:

A1=1, A2=2, A3=4;

A1=1, A2=6;

A1=2, A2=5;

A1=3, A2=4.

样例输入

7 2

样例输出

4

题意:给一个数N，能分解成多少个数，这些数满足题目所说的4个条件。问有多少种情况？
思路：看到N最大只有100,1+2+3...+14>100最多14层递归。用DFS找出所有情况。
感想：1.其中有个乘法要注意下数据范围，第一用了int，WA了，后来改用long long。
     2.题目数据貌似水了，这题要求GCD，目前还没看懂，先贴个水过的代码，以后再补上。

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
using namespace std;

const int INF=0x3f3f3f3f;
const double eps=1e-10;
const double PI=acos(-1.0);
#define maxn 500
int vis[maxn],a[maxn];
int n, m, cnt;
void dfs(int sum, long long mul, int pos, int num)
{
    if(sum == n && mul%m ==0)
    {
//        for(int i = 0; i < num; i++)
//            printf("%d ", a[i]);
//        printf("
");
        cnt++;
        return;
    }

    for(int i = pos; i <= n; i++)
    {
        if(sum + i > n)
            break;
        a[num] = i;
        dfs(sum+i, mul*i, i+1, num+1);
    }
}
int main()
{
    scanf("%d%d", &n, &m);
    cnt = 0;
    dfs(0,1,1,0);
    printf("%d
", cnt%1000000007);
    return 0;
}