思路：
根据题意可知A、B、C、D四个人只有1个做了好事，则有以下4种状态：（做好事的为1，其他的为0）
状态1：A == 1 B == 0 C == 0 D == 0
状态2：A == 0 B == 1 C == 0 D == 0
状态3：A == 0 B == 0 C == 1 D == 0
状态4：A == 0 B == 0 C == 0 D == 1
所谓枚举，就是按照这4种状态的假定，去逐一的去测试4个人中有几句真话
若有3句则该状态成立，其他则该状态不成立

#include <iostream>

using namespace std;

int main()
{
    for(int a = 0; a <= 1; a++)
    {
        for(int b = 0; b <= 1; b++)
        {
            for(int c = 0; c <= 1; c++)
            {
                for(int d = 0; d <= 1; d++)
                {
                    if(a + b + c + d == 1) // 限定做好事的人的个数
                    {
                        int cc1 = (a == 0);
                        int cc2 = (c == 1);
                        int cc3 = (d == 1);
                        int cc4 = (d == 0);

                        if(cc1 + cc2 + cc3 + cc4 == 3)
                        {
                            cout << "A" << (a == 1 ? "做了" : "没做") << "好事" << endl;
                            cout << "B" << (b == 1 ? "做了" : "没做") << "好事" << endl;
                            cout << "C" << (c == 1 ? "做了" : "没做") << "好事" << endl;
                            cout << "D" << (d == 1 ? "做了" : "没做") << "好事" << endl;
                        }
                    }
                }
            }
        }
    }

    return 0;
}