Topic 1125: - delegate * C language training

题目描述

某侦察队接到一项紧急任务,要求在A、B、C、D、E、F六个队员中尽可能多地挑若干人,但有以下限制条件：
1)A和B两人中至少去一人;
2)A和D不能一起去;
3)A、E和F三人中要派两人去;
4)B和C都去或都不去;
5)C和D两人中去一个;
6)若D不去,则E也不去.
问应当让哪几个人去？

输入格式

无

输出格式

person to send
若有多个,In ascending alphabetical order,用逗号分开（with trailing comma）
 

样例输入

复制

无

样例输出

复制

A,B,C,F,

This is my first time doing this shit,,No clue at first,Still want to draw lessons from bosses

1.思路,创建变量,初始值为0 ,,逐步增加,1代表去,0 代表不去

然后用for循环遍历,,The previous code that can be passed

#include<stdio.h>
void print(int a, int b, int c, int d, int e, int f)
{
	if (a == 1)
	{
		printf("A,");
	}
	if (b == 1)
	{
		printf("B,");
	}
	if (c == 1)
	{
		printf("C,");
	}
	if (d == 1)
	{
		printf("D,");
	}
	if (e == 1)
	{
		printf("E,");
	}
	if (f == 1)
	{
		printf("F,");
	}
}

int judg(int a, int b, int c, int d, int e, int f)
//Here I found the opposite of each request,,,ideas than in aifThere is a lot of writing in it to be clear
{
	if (a == 0 && b == 0)
	{
		return 0;
	}
	if (a == 1 && d == 1)
	{
		return 0;
	}
	if ((b == 0 && c == 1) || (b == 1 && c == 0))
	{
		return 0;
	}
	if ((c == 0 && d == 0) || (c == 1 && d == 1))
	{
		return 0;
	}
	if (d == 0 && e == 1)
	{
		return 0;
	}

	if (a == 0 && e == 0 && f == 0)//去0个
	{
		return 0;
	}
	if ((a == 1 && e == 0 && f == 0) || (a == 0 && e == 1 && f == 0) || (a == 0 && e == 0 && f == 1))
	{
		return 0;
	}
	if (a == 1 && e == 1 && f == 1)
	{
		return 0;
	}

	return 1;
}
int main()
{
	int a = 0; int b = 0; int c = 0;
	int d = 0; int e = 0; int f = 0;

	for (a = 0; a < 2; a++)
	{
		for (b = 0; b < 2; b++)
		{
			for (c = 0; c < 2; c++)
			{
				for (d = 0; d < 2; d++)
				{
					for (e = 0; e < 2; e++)
					{
						for (f = 0; f < 2; f++)
						{
							int k = judg(a, b, c, d, e, f);//判断是否满足条件
							if (k == 1)
							{
								print(a, b, c, d, e, f);//输出函数
								return 0;
							}
						}
					}
				}
			}
		}
	}

	return 0;

}

 Then let's improve,The problem with this code is

1. Pass if you meet the conditions,无法找到最优解

2.Difficulty in passing ginseng,,To several parameters,,懒！

3.There are also different methods in the judgment process

改进：对于创建6个变量,Why not create an array,Then record each possible value,Finally find meet the conditions of the optimal solution！！！@#￥%……

#include<stdio.h>
void change(int arr[10], int a, int b, int c, int d, int e, int f)
{
		if (a == 1)
	{
			arr['a' - 'a' + 1]++;
	}
	if (b == 1)
	{
		arr['b' - 'a' + 1]++;
	}
	if (c == 1)
	{
		arr['c' - 'a' + 1]++;
	}
	if (d == 1)
	{
		arr['d' - 'a' + 1]++;
	}
	if (e == 1)
	{
		arr['e' - 'a' + 1]++;
	}
	if (f == 1)
	{
		arr['f' - 'a' + 1]++;
	}
}
void print(int arr[10])
{
	int i = 0;
	for (; i<7; i++)
	{
		if (arr[i] == 1)
		{
			printf("%c,", i + 'A' - 1);
		}
	}
}

int judg(int a, int b, int c, int d, int e, int f)
{
	if (a == 0 && b == 0)
	{
		return 0;
	}
	if (a == 1 && d == 1)
	{
		return 0;
	}
	if ((b == 0 && c == 1) || (b == 1 && c == 0))
	{
		return 0;
	}
	if ((c == 0 && d == 0) || (c == 1 && d == 1))
	{
		return 0;
	}
	if (d == 0 && e == 1)
	{
		return 0;
	}

	if (a == 0 && e == 0 && f == 0)//去0个
	{
		return 0;
	}
	if ((a == 1 && e == 0 && f == 0) || (a == 0 && e == 1 && f == 0) || (a == 0 && e == 0 && f == 1))
	{
		return 0;
	}
	if (a == 1 && e == 1 && f == 1)
	{
		return 0;
	}

	int num = 0;
	if (a == 1) { num++; }
	if (b== 1) { num++; }
	if (c  == 1) { num++; }
	if (d== 1) { num++; }
	if (e  == 1) { num++; }
	if (f== 1) { num++; }
	return num;
}
int main()
{

	//创建一个只有7位的数组,用123来代替abc,
	//
	// Then in order to record,I thought of using a two-dimensional array,或者结构体,
	//有2的6method in power,Just two of the six to the power line

	int i = 0;


	int a = 0; int b = 0; int c = 0;
	int d = 0; int e = 0; int f = 0;
	int arr[100][10] = { 0 };
	
		for (a = 0; a < 2; a++)
		{
			for (b = 0; b < 2; b++)
			{
				for (c = 0; c < 2; c++)
				{
					for (d = 0; d < 2; d++)
					{
						for (e = 0; e < 2; e++)
						{
							for (f = 0; f < 2; f++)
							{
								int k = judg(a, b, c, d, e, f);
								if (k!=0)
								{
									arr[i][0] = k;
									change(arr[i], a, b, c, d, e, f);
									i++;
								}
							}
						}
					}
				}
			}
		}
	//

		int max = arr[0][0];
		int max_id = 0;
		int j = 0;
		for (j = 0; j < i; j++)
		{
			if (arr[j][0] > max)
			{
				max = arr[j][0];
				max_id = j;
			}
		}

		print(arr[max_id]);
		return 0;
}

尽力了,In this way, it is possible to find the optimal solution