吃豆子卡牌组合技公平的匹配赛_动态规划吃豆子-优快云博客

本文解析了三道使用C++实现的经典编程题目：吃豆子游戏的路径计算、卡牌组合技的价值最大化及公平匹配赛的可行性判断。通过动态规划等算法展示了问题解决的思路。

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忘记题目了，吃豆子游戏

#include<math.h>
#include<stdio.h>
#include<string.h>
#include<assert.h>
#include<limits.h>
#include<stdbool.h>
#include<stdlib.h>
#include<iostream>
using namespace std;
 
int  data[150][150] = { 0 };
 
 

 

int main()
{
	 
		
		int n, m;
		while (cin >> n >> m)
		{
			memset(data, 0, 150 * 150);
			for (int i = 0; i < n; i++)
			{
				for (int j = 0; j < m; j++)
				{
					char c;
					cin >> c;
					if ('W' == c)
						data[i][j] = 1;
				}
			}

			int ii;
			int flag = true;
			for (ii = n-1; ii>=0 && flag; ii--)
			{
				for (int j = 0; j < m && flag; j++)
				{
					 
					if (data[ii][j]> 0 && flag )
					{
						flag = false;
					}
				}
			 
			}
			ii++;
			int count = 0;
			int x = 0;
			int y = -1; 
			int r = -1;
			for (int i = 0; i <= ii; i++)
			{
				int begin = 0;
				int end = m - 1;
				while (begin <n && data[i][begin] == 0 )
					begin++;
				while (end >= 0 && data[i][end] == 0)
					end--;

				if (begin == n && end == -1)
				{
					y++;
					r = -r;
					count++;
				}

				else if (begin <= end)
				{
					if (r == -1)
					{
						if (x <= begin)
						{
							y++;
							r = -r;
						 
							count += end - x + 1;
							x = end;
						}
						else
						{
							y++;
							r = -r;
							count += end - begin + x - begin + 1;
							x = end;
						}
					}
					else
					{
						if (x >= end)
						{
							y++;
							r = -r;

							count += x - begin+ 1;
							x = begin;
						}
						else
						{
							y++;
							r = -r;
							count += end - begin + end - x + 1;
							x = end;
						}
					}


				}





			}

			


			cout << count - 1;


		}

	 

}

卡牌组合技

#include<math.h>
#include<stdio.h>
#include<string.h>
#include<assert.h>
#include<limits.h>
#include<stdbool.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;
 
int  data[20]  = { 0 };
int da[20][10000007] = { 0 };
int sum = 0;
 
 

 

int main()
{
	 
		
		int n, m;
		while (cin >> n >> m)
		{
			for (int i = 0; i < n; i++)
			{
				cin >> data[i];
				sum += data[i];
			}
			std::sort(data, data + n);
			int big = -1000000007;

			memset(da, 0, sum + 1);
			for (int i = 0; i < n; i++)
			{
				da[i][0] = 1;
			}

			da[0][data[0]] = data[0];
		
			for (int i = 1; i < n; i++)
			{
				for (int j = 0; j < data[i]; j++)
				{

					da[i][j] = da[i - 1][j];
					
				}

				for (int j = data[i] ; j <= m; j++)
				{
					da[i][j] = std::max(da[i - 1][j], da[i - 1][j - data[i]] * data[i]);
				}

					 

			}


			cout << da[n - 1][m] << endl;;

          

			 

		}



	 

}

公平的匹配赛

#include<math.h>
#include<stdio.h>
#include<string.h>
#include<assert.h>
#include<limits.h>
#include<stdbool.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;
int data[100];
int da[100][10000] = { 0};
int sum = 0;
 
bool equit(int* data, int n)
{
	if (n % 2)
		return false;
	int sum = 0;
	for (int i = 0; i < n; i++)
	{
		sum += data[i];
	}
	
	da[0][data[0]] = 1;

	for (int i = 1; i < n; i++)
	{
		for (int j = 0; j < data[i]; j++)
		{
			da[i][j] = da[i - 1][j];
		}

		for (int j = data[i]; j <= sum / 2; j++)
		{
			da[i][j] = std::max(da[i-1][j - data[i]] + 1, da[i-1][j]);
		}
	}


	if (da[n - 1][sum / 2] >= n / 2)
		return true;

	else
		return false;



}

 

int main()
{
	 
		
		int n, m;
		cin >> n >> m;
		{
			memset(data, 0, 100);
			memset(da, 0, 10000 * 100);
			for (int j = 0; j < n; j++)
			{
				{
					int sum = 0;
					bool result = false;
					for (int i = 0; i < m; i++)
					{
						cin >> data[i];

					}

					result = equit(data, m);
					if (result)
						cout << "TRUE" << endl;
					else
						cout << "FALSE" << endl;






				}
			}

		}



	 

}

这次犯了错误，for循环里面for循环break只对里层for循环有效，切记切记。后面两道动态规划的一时没想出来，交完卷一个小时就写出来了，也没有在线编译，例子是通过的