2017年上海金马五校程序设计竞赛 C : Count the Number 深搜

Description

Given n numbers, your task is to insert '+' or '-' in front of each number to construct expressions. Note that the position of numbers can be also changed.

You can calculate a result for each expression. Please count the number of distinct results and output it.

 

Input

There are several cases.
For each test case, the first line contains an integer n (1 ≤ n ≤ 20), and the second line contains n integers a₁,a₂, ... ,a_n(-1,000,000,000 ≤ a_i ≤ 1,000,000,000).

 

Output

For each test case, output one line with the number of distinct results.

 

Sample Input

21 231 3 5

 

Sample Output

48

其实就是一个递归啦，别想太多啦，迷迷糊糊就能过

#include <iostream>#include <set>using namespace std;int a[25];set <int> q;void getsum(int sum,int n,int m){if(n == m){q.insert(sum);return;}getsum(sum+a[n],n+1,m);getsum(sum-a[n],n+1,m);}int main(){int n;while(scanf("%d",&n)!=EOF){for(int i = 0;i < n; i++){scanf("%d",&a[i]);}q.clear();getsum(0,0,n);int u = q.size();printf("%d\n",u);}}