SPOJ AMR10I Dividing Stones

Time limit: 7s Source limit: 50000B Memory limit: 256MB

The first line contains the number of test cases T. T lines follow, one corresponding to each test case, containing
2 integers: N and P. 
 
OUTPUT

3 
6 
 
EXPLANATION
In the first test case, the possible ways of division are (1,1,1), (1,2), (2,1) and (3) which have values
1, 2, 2, 3 and hence, there are 3 distinct values. 
In the second test case, the numbers 1 to 6 constitute the answer and they can be obtained in the following
ways: 
1=1*1*1*1*1 
2=2*1*1*1 
3=3*1*1 
4=4*1 
5=5 
6=2*3

题意:有n个石子，可以分成任意堆，每一种分法的值为每一堆的石子数量的乘积。求一共可以分成多少个不同的乘积。

分析：最终的乘积除了1以外，都可以分解成素数相乘或者素数相乘再与1相乘的形式。因为n不超过70，所以我们可以先找出不超过70的所有素数，然后从这些素数中进行搜索求解即可。为了方便求出不同的乘积有多少个，可以用STL里面的set来统计不同的数有多少个。

 

 

 

#include<cstdio>
#include<cstring>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
set<LL> s;
int prime[] = {, , , , , , , , , , , , , , , , , , , , };
int n, p;

void dfs(int num, int cur, LL ans)
{
    s.insert(ans);
    if(cur < prime[num]) return ;
    dfs(num, cur - prime[num], ans * prime[num] % p); //要第num个素数
    dfs(num+, cur, ans); //不要第num个素数
}

int main()
{
    int T, i, j;
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&n,&p);
        s.clear();
        dfs(, n, );
        printf("%d\n", s.size());
    }
    printf("\n");
}