【codeforce508C】. Anya and Ghosts

本文探讨了一个有趣的问题:如何计算在特定条件下所需的最小数量的蜡烛,以确保每次鬼来访时都有足够的蜡烛点亮。文章详细解释了输入参数的意义,并通过示例展示了算法的实现过程。

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C. Anya and Ghosts
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Anya loves to watch horror movies. In the best traditions of horror, she will be visited by m ghosts tonight. Anya has lots of candles prepared for the visits, each candle can produce light for exactly t seconds. It takes the girl one second to light one candle. More formally, Anya can spend one second to light one candle, then this candle burns for exactly t seconds and then goes out and can no longer be used.

For each of the m ghosts Anya knows the time at which it comes: the i-th visit will happen wi seconds after midnight, all wi's are distinct. Each visit lasts exactly one second.

What is the minimum number of candles Anya should use so that during each visit, at least r candles are burning? Anya can start to light a candle at any time that is integer number of seconds from midnight, possibly, at the time before midnight. That means, she can start to light a candle integer number of seconds before midnight or integer number of seconds after a midnight, or in other words in any integer moment of time.

Input

The first line contains three integers m, t, r (1 ≤ m, t, r ≤ 300), representing the number of ghosts to visit Anya, the duration of a candle's burning and the minimum number of candles that should burn during each visit.

The next line contains m space-separated numbers wi (1 ≤ i ≤ m, 1 ≤ wi ≤ 300), the i-th of them repesents at what second after the midnight the i-th ghost will come. All wi's are distinct, they follow in the strictly increasing order.

Output

If it is possible to make at least r candles burn during each visit, then print the minimum number of candles that Anya needs to light for that.

If that is impossible, print  - 1.

Examples
Input
1 8 3
10
Output
3
Input
2 10 1
5 8
Output
1
Input
1 1 3
10
Output
-1
Note

Anya can start lighting a candle in the same second with ghost visit. But this candle isn't counted as burning at this visit.

It takes exactly one second to light up a candle and only after that second this candle is considered burning; it means that if Anya starts lighting candle at moment x, candle is buring from second x + 1 to second x + t inclusively.

In the first sample test three candles are enough. For example, Anya can start lighting them at the 3-rd, 5-th and 7-th seconds after the midnight.

In the second sample test one candle is enough. For example, Anya can start lighting it one second before the midnight.

In the third sample test the answer is  - 1, since during each second at most one candle can burn but Anya needs three candles to light up the room at the moment when the ghost comes.

一个人要遇到m个鬼,遇到每个鬼的时间为w[i],蜡烛亮着的持续时间为t,遇到一个鬼至少要点燃r支蜡烛,
//点燃的时间不算燃烧时间,求他最少要带多少蜡烛 ,不存在就输出-1

code:

#include<cstdio>
#include<algorithm>
using namespace std;
#include<cstring> 
int num[1000];//某一时刻有多少蜡烛亮着
int w[1000];//每个鬼到的时间 
int main()
{
	int m,r,t;
	while(~scanf("%d%d%d",&m,&t,&r))
	{
		for(int i=1;i<=m;i++)
			scanf("%d",&w[i]);
			if(r>t){//持续时间比蜡烛数少,蜡烛未全部点燃,有的已经熄灭 
				printf("-1\n");
				continue;
			}
			memset(num,0,sizeof(num));
			int ans=0;//所需蜡烛数
			for(int i=1;i<=m;i++)//拜访每一个鬼 
			{
				for(int j=w[i]-1;num[w[i]]<r;j--){//从鬼来时的前一秒开始倒着点蜡烛 
				for(int k=j+1;k<=j+t;k++){//每个蜡烛持续时间段 
					if(k>=0)//j可能为0 
					 num[k]++; 
				}
				ans++;	
				} 
			 } 
			 printf("%d\n",ans);
	 } 
	return 0;
 } 


### 回答1: #include <stdio.h> int main(){ //定义一个3*3的数组用来存储棋盘 int board[3][3]={0}; int x,y,turn=1; while (1){ //打印棋盘 printf("当前棋盘:\n"); for (x=0;x<3;x++){ for (y=0;y<3;y++){ printf("%d ",board[x][y]); } printf("\n"); } //根据turn的值来判断谁轮到落子 if (turn==1){ printf("轮到X落子,请输入落子的位置(x y):"); }else { printf("轮到O落子,请输入落子的位置(x y):"); } scanf("%d %d",&x,&y); //将落子位置的值设置为对应的值 board[x][y] = turn; //改变轮到谁落子 turn = -turn; //判断谁赢了 if (board[0][0]==board[1][1] && board[1][1]==board[2][2] && board[2][2]!=0){ printf("游戏结束,获胜者是%c\n",board[0][0]==1?'X':'O'); break; } if (board[2][0]==board[1][1] && board[1][1]==board[0][2] && board[0][2]!=0){ printf("游戏结束,获胜者是%c\n",board[2][0]==1?'X':'O'); break; } for (x=0;x<3;x++){ if (board[x][0]==board[x][1] && board[x][1]==board[x][2] && board[x][2]!=0){ printf("游戏结束,获胜者是%c\n", board[x][0] == 1 ? 'X' : 'O'); break; } if (board[0][x]==board[1][x] && board[1][x]==board[2][x] && board[2][x]!=0){ printf("游戏结束,获胜者是%c\n", board[0][x] == 1 ? 'X' : 'O'); break; } } } return 0; } ### 回答2: 为了回答这个问题,需要提供题目的具体要求和规则。由于提供的信息不够具体,无法为您提供准确的代码。但是,我可以给您一个简单的Tic-tac-toe游戏的示例代码,供您参考: ```c #include <stdio.h> #include <stdbool.h> // 判断游戏是否结束 bool isGameOver(char board[][3]) { // 判断每行是否有3个相同的棋子 for(int i = 0; i < 3; i++) { if(board[i][0] != '.' && board[i][0] == board[i][1] && board[i][0] == board[i][2]) { return true; } } // 判断每列是否有3个相同的棋子 for(int i = 0; i < 3; i++) { if(board[0][i] != '.' && board[0][i] == board[1][i] && board[0][i] == board[2][i]) { return true; } } // 判断对角线是否有3个相同的棋子 if(board[0][0] != '.' && board[0][0] == board[1][1] && board[0][0] == board[2][2]) { return true; } if(board[0][2] != '.' && board[0][2] == board[1][1] && board[0][2] == board[2][0]) { return true; } return false; } // 输出棋盘 void printBoard(char board[][3]) { for(int i = 0; i < 3; i++) { for(int j = 0; j < 3; j++) { printf("%c ", board[i][j]); } printf("\n"); } } int main() { char board[3][3]; // 初始化棋盘 for(int i = 0; i < 3; i++) { for(int j = 0; j < 3; j++) { board[i][j] = '.'; } } int player = 1; // 玩家1先下 int row, col; while(true) { printf("Player %d's turn:\n", player); printf("Row: "); scanf("%d", &row); printf("Column: "); scanf("%d", &col); // 判断输入是否合法 if(row < 0 || row >= 3 || col < 0 || col >= 3 || board[row][col] != '.') { printf("Invalid move. Try again.\n"); continue; } // 下棋 board[row][col] = (player == 1) ? 'X' : 'O'; // 输出棋盘 printBoard(board); // 判断游戏是否结束 if(isGameOver(board)) { printf("Player %d wins!\n", player); break; } // 切换玩家 player = (player == 1) ? 2 : 1; } return 0; } ``` 这段代码实现了一个简单的命令行下的Tic-tac-toe游戏。玩家1使用'X'棋子,玩家2使用'O'棋子。玩家依次输入行和列,下棋后更新棋盘,并判断游戏是否结束。当游戏结束时,会输出获胜者并结束游戏。 ### 回答3: 题目要求实现一个井字棋游戏的判断胜负函数。给定一个3x3的井字棋棋盘,用C语言编写一个函数,判断当前是否存在某个玩家获胜或者平局。 题目要求代码中定义一个3x3的字符数组board来表示棋盘,其中 'X' 表示玩家1在该位置放置了一个棋子, 'O' 表示玩家2在该位置放置了一个棋子, '.' 表示该位置没有棋子。 下面是实现此题的C语言代码: ```c #include <stdio.h> #include <stdbool.h> // 用于使用bool类型 bool checkWin(char board[3][3]) { // 检查每一行是否有获胜的情况 for (int row = 0; row < 3; row++) { if (board[row][0] == board[row][1] && board[row][1] == board[row][2] && board[row][0] != '.') { return true; } } // 检查每一列是否有获胜的情况 for (int col = 0; col < 3; col++) { if (board[0][col] == board[1][col] && board[1][col] == board[2][col] && board[0][col] != '.') { return true; } } // 检查对角线是否有获胜的情况 if ((board[0][0] == board[1][1] && board[1][1] == board[2][2] && board[0][0] != '.') || (board[0][2] == board[1][1] && board[1][1] == board[2][0] && board[0][2] != '.')) { return true; } return false; // 没有获胜的情况 } int main() { char board[3][3]; // 存储棋盘状态 // 读取棋盘状态 for (int i = 0; i < 3; i++) { scanf("%s", board[i]); } // 调用检查胜负的函数,并输出结果 if (checkWin(board)) { printf("YES\n"); } else { printf("NO\n"); } return 0; } ``` 这个程序中定义了一个函数checkWin,用于检查是否有玩家获胜。遍历棋盘的每一行、每一列和对角线,判断是否有连续相同的字符且不为'.',如果有,则返回true;否则返回false。 在主函数main中,首先定义一个3x3的字符数组board,然后通过循环从标准输入中读取棋盘状态。接着调用checkWin函数进行胜负判断,并根据结果输出"YES"或者"NO"。最后返回0表示程序正常结束。 请注意,该代码只包含了检查胜负的功能,并没有包含其他如用户输入、判断平局等功能。如果需要完整的游戏代码,请告知具体要求。
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