Problem C : Count the Number

Time Limit: 3 s

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, … ,an (-1,000,000,000 ≤ ai ≤ 1,000,000,000).

Output

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

Sample Input

2
1 2
3
1 3 5

Sample Output

4
8

分析

題意：給一組數，每個數字前需要加一個“+”或者一個“-”，最後將全部數求和，計算出各種情況的下總和的值不同的數目。很簡單，時間限制3s，所以直接暴力

程式碼

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cstdlib>
 

#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
int a[22],b[2000000];
int N,sum,sumn1;
void counth(int i,queue<int>q)
{
    if(i>=N)///全部數都加上之後計算不同的和的數量
    {
        int zsum;
        zsum=q.size();
        sum=zsum;
        for(int k=0;k<zsum;k++)
        {
            b[k]=q.front();
            q.pop();
        }///將佇列中的元素存入陣列中
        sort(b,b+zsum);///將陣列元素排序用來刪重
        for(int k=1;k<zsum;k++)
            if(b[k]==b[k-1])sum--;
        return;
    }
    sumn1=q.size();
    for(int j=0;j<sumn1;j++)
    {
        q.push(q.front()+a[i]);///將各個數字的兩種情況都入隊
        q.push(q.front()-a[i]);
        q.pop();///將計算過的總和出隊
    }
    i++;
    counth(i,q);
}
int main()
{
    while(~scanf("%d",&N))
    {
        queue<int>q;///定義佇列存放不同情況下的和
        for(int k=0;k<N;k++)
            scanf("%d",&a[k]);
        q.push(a[0]);
        q.push(-a[0]);///將第一個數入隊
        counth(1,q);
        printf("%d\n",sum);
    }
}