#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int res = 0;//全局变量返回值void dfs(int sum, int mult, int index, bool flag, vector<int> & num){//flag是用来标记与前一个数字重复时前一个数字是否被计入。若前一个被计入则当前节点可计入，也可不计入；若前一节点未被计入则当前节点也不计入。（为了防止重复。）if(index == num.size()) return; //边界处理int newsum = sum+num[index];int newmult = mult*num[index];if(index == 0){　　//第一个数的时候特殊处理，若最小的数不是1直接返回0if(num[0] == 1) {dfs(newsum, newmult,index+1,true,num);dfs(sum, mult,index+1,false,num);}else{return ;}}else{if(num[index] == num[index-1]){//与前一节点重复if(flag){//true的时候当前节点分两种讨论if(newsum>newmult) {//当前节点加入是否符合要求，若符合则继续，否则直接退出（剪枝）res+=1;dfs(newsum, newmult,index+1,true,num);}dfs(sum, mult,index+1,false,num);}else{//重复且前一节点未计入，则当前也不计入dfs(sum, mult,index+1,false,num);}}else{if(newsum>newmult) {res+=1;dfs(newsum, newmult,index+1,true,num);}dfs(sum, mult,index+1,false,num);}}
}int main(){int counts = 0,temp;vector<int> nums;cin>>counts;for(int i=0;i<counts;++i){cin>>temp;nums.push_back(temp);}sort(nums.begin(),nums.end());//排序有利于后续剪枝dfs(0,1,0,false,nums);cout<<res;return 0;
}

#include <iostream>
#include <stdlib.h>
using namespace std;int n;
int nums[1000];int cmp(const void * a, const void * b) {return *(int*)a - *(int*)b;
}// 思路：DFS生成全组合，同时注意剪枝、避免重复组合
int findall(int nums[], int index, long sum, long multi) {int count = 0;for(int i=index; i<n; i++) {sum += nums[i];multi *= nums[i];if(sum > multi)count += 1 + findall(nums, i+1, sum, multi);else if(nums[i] == 1)count += findall(nums, i+1, sum, multi);elsebreak;sum -= nums[i];multi /= nums[i];// 跳过相等的元素，避免重复组合while(i<n-1 && nums[i]==nums[i+1])i++;}return count;
}int main(int argc, char* argv[])
{while(cin >> n) {for(int i=0; i<n; i++)cin >> nums[i];// 从小到大排序qsort(nums, n, sizeof(int), cmp);cout << findall(nums, 0, 0, 1) << endl;}return 0;
}