C语言实例：搬山游戏

*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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