编程挑战：商人过河逻辑游戏

编程完成商人过河游戏：有三个商人带着三个随从和货物过河，船每次最多只能载两个人，由他们自己划行，并且如何乘船渡河的大权由商人掌握。要求保证在过河期间的任一岸上商人的人数要大于或等于随从的人数，否则随从会杀死商人抢走货物。设计一个符合上述要求的商人过河的游戏



#include<stdio.h>
#include<stdlib.h>

void print()                     //游戏屏幕
{
   system("cls");
   printf("****************************************************************\n");
   printf("*                                                              *\n");
   printf("*              ^_^welcome to the game!^_^                      *\n");
   printf("*                                                              *\n");
   printf("*   Game rules:                                                *\n");
   printf("*    3 men and 3 rateiners and goods to pass the river,the boat*\n");
   printf("*  carrys 2 persons each time.In passing the river,at any bank *\n");
   printf("*  number of men must be more than the number the rateiners ,  *\n");
   printf("*  or the rateiners will kill the men and take the goods.      *\n");
   printf("*  Game operations:  1.Input the number of men and rateiners   *\n");
   printf("*                       in turn.                               *\n");
   printf("*                    2.Input error keys,the game will restart. *\n");
   printf("****************************************************************\n");
}

void began()                //游戏开始界面
 {
 char ch;
 printf("\n\n");
 printf("      Press any key to start the game.(Q key to quit)...");

 scanf("%c",&ch);
 if(ch=='Q'||ch=='q')
 exit(0);
 }
void xianshi(char *a,char *b)    //  显示过河的状态
{
  int ax=0,ay=0,bx=0,by=0;        //ax，ay,bx,by分别表示岸两边的商人和仆人数
  int i,j;
  for(i=0;i<6;i++)
  {
  if(*(a+i)=='M')
  ax++;
  if(*(a+i)=='S')
  ay++;
  if(*(b+i)=='M')
  bx++;
  if(*(b+i)=='S')
  by++;
  }
  printf("This bank\n");
  for(i=1;i<=ax;i++)
  printf("man ");
  printf("\n");
  for(i=1;i<=ay;i++)
  printf("rateiner ");
  printf("\n\n");
  printf("That bank\n");
  for(i=1;i<=bx;i++)
  printf("man ");
   printf("\n");