I - Dividing Stones

Description

There are N stones, which can be divided into some piles arbitrarily. Let the value of each division be equal to the product of the number of stones in all the piles modulo P. How many possible distinct values are possible for a given N and P?
 
INPUT
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
Output T lines, each line containing the required answer for the corresponding test case.
 
CONSTRAINTS
T <= 20
2 <= N <= 70
2 <= P <= 1e9
 
SAMPLE INPUT
2
3 1000
5 1000
 
SAMPLE 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：n  p    在1~~n内 寻找素数和等于n的  并且乘积不相等

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<set>
#include <iostream>

using namespace std;
set<long>s;
int p[80],mod,n,t;
bool vis[80];

void prime()
{
    int cnt=0;
    memset(vis,1,sizeof(vis));
    memset(p,0,sizeof(p));
    for(int i=2; i<=80; i++)  ///素数打标   n很小 <=70  所以可以打表
    {
        if(vis[i])
        {
            p[++cnt]=i;
            for(int j=i*i; j<=80; j+=i)
                vis[j]=0;
               // cout<<p[cnt-1]<<"~~~~~~~~~~~"<<endl;
        }

    }
}
void dfs(int i,int x,long long ans)
{
    s.insert(ans);
    if (p[i]>x)
        return;
    dfs(i,x-p[i],ans*p[i]%mod);  ///取第i个素数
    dfs(i+1,x,ans);  ///深搜第i+1个素数
}
int main()
{
    prime();
    scanf("%d",&t);
    while (t--)
    {
        s.clear();
        scanf("%d%d",&n,&mod);
        dfs(1,n,1);
        printf("%d\n",s.size());
    }
    return 0;
}
//dfs(x,n-pr[x],ji*pr[x]%p);//取第x个素数
//dfs(x+1,n,ji);//从第x+1个素数深搜

I - Dividing Stones