C语言实例:搬山游戏

来源:网络时间:2011-06-24

  设有n座山,计算机与人为比赛的双方,轮流搬山。规定每次搬山的数止不能超 过k座,谁搬最后一座谁输。游戏开始时。计算机请人输入山的总数(n)和每次允许搬山的最大数止(k)。然后请人开始,等人输入了需要搬走的山的数目后,计算机马上打印出它搬多少座山,并提示尚余多少座山。双方轮流搬山直到最后一座山搬完为止。计算机会显示谁是赢家,并问人是否要继续比赛。若人不想玩了,计算机便会统计出共玩了几局,双方胜负如何。

  *问题分析与算法设计

  计算机参加游戏时应遵循下列原则:

  1) 当:

  剩余山数目-1<=可移动的最大数k 时计算机要移(剩余山数目-1)座,以便将最后一座山留给人。

  2)对于任意正整数x,y,一定有:

  0<=x%(y+1)<=y

  在有n座山的情况下,计算机为了将最后一座山留给人,而且又要控制每次搬山的数目不超过最大数k,它应搬山的数目要满足下列关系:

  (n-1)%(k+1)

  如果算出结果为0,即整除无余数,则规定只搬1座山,以防止冒进后发生问题。

  按照这样的规律,可编写出游戏程序如下:

  #include

  int main()

  {

  int n,k,x,y,cc,pc,g;

  printf("More Mountain Game\n");

  printf("Game Begin\n");

  pc=cc=0;

  g=1;

  for(;;)

  {

  printf("No.%2d game \n",g++);

  printf("---------------------------------------\n");

  printf("How many mpuntains are there?");

  scanf("%d",&n);

  if(!n) break;

  printf("How many mountains are allowed to each time?");

  do{

  scanf("%d",&k);

  if(k>n||k<1) printf("Repeat again!\n");

  }while(k>n||k<1);

  do{

  printf("How many mountains do you wish movw away?");

  scanf("%d",&x);

  if(x<1||x>k||x>n) /*判断搬山数是否符合要求*/

  {

  printf("IIIegal,again please!\n");

  continue;

  }

  n-=x;

  printf("There are %d mountains left now.\n",n);

  if(!n)

  {

  printf("...............I win. You are failure...............\n\n");cc++;

  }

  else

  {

  y=(n-1)%(k+1); /*求出最佳搬山数*/

  if(!y) y=1;

  n-=y;

  printf("Copmputer move %d mountains away.\n",y);

  if(n) printf(" There are %d mountains left now.\n",n);

  else

  {

  printf("...............I am failure. You win..................\n\n");

  pc++;

  }

  }

  }while(n);

  }

  printf("Games in total have been played %d.\n",cc+pc);

  printf("You score is win %d,lose %d.\n",pc,cc);

  printf("My score is win %d,lose %d.\n",cc,pc);

  }

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最新评论(共1条)

2345市网友

游戏的智能方面略差,用户友好度也不够啊

2013-02-14 0

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