三国杀张昌蒲计算器

三国杀 张昌蒲 严教 计算器

基于https://blog.youkuaiyun.com/qq_45716735/article/details/124362234更改，修复了一些bug
输入点数，中间用空格隔开，按下回车键结束输入

#include <bits/stdc++.h>
using namespace std;
set<string> Ans;
int input[100];
int i;
int to;
void work(int *a, int len)
{
    int sum1 = 0;
    int sum2 = 0;
    for (int mid = 1; mid < len; mid++)
    {
        string tempAns1 = "", tempAns2 = "";
        int temp[len];
        sum2 = 0;
        sum1 = 0;
        for (int k = 0; k < mid; k++)
        {
            sum1 += a[k];
            temp[k] = a[k];
        }
        sort(temp, temp + mid);
        for (int k = 0; k < mid; k++)
        {
            tempAns1 += to_string(temp[k]) + " ";
        }
        for (int k = mid; k < len; k++)
        {
            sum2 += a[k];
            temp[k] = a[k];
        }
        sort(temp + mid, temp + len);
        for (int k = mid; k < len; k++)
        {
            tempAns2 += to_string(temp[k]) + " ";
        }
        if (sum1 == sum2)
        {
            string temps = tempAns1;
            tempAns1 += tempAns2;
            tempAns2 += temps;
            if (Ans.count(tempAns1) == 0 && Ans.count(tempAns2) == 0)
            {
                Ans.insert(tempAns1);
                cout << "情况" << to++ << ":" << endl;
                for (int k = 0; k < mid; k++)
                {
                    cout << a[k] << " ";
                }
                cout << endl;
                for (int k = mid; k < len; k++)
                {
                    cout << a[k] << " ";
                }
                cout << endl
                     << "总分牌数：" << len << "     总和：" << sum1 << endl
                     << endl;
            }
        }
        tempAns1.clear();
        tempAns2.clear();
    }
}

int main()
{
    to = 1;
    string s;
    getline(cin, s);
    istringstream iss(s);
    string ans;
    while (iss >> ans)
    {
        input[i++] = stoi(ans);
    }
    sort(input, input + i);
    for (int len = i; len > 0; len--)
    {
        Ans.clear();
        work(input, len);
        while (next_permutation(input, input + i)) //对input数组前i个数进行全排列,选中部分数组会逐渐从大到小排列，当不存在下一个全排列后会将数组从小到大排列
        {
            work(input, len);
        }
    }
    system("pause");
    return 0;
}