089.配对新郎和新娘

  

   

    

     

    
    

     

      #include<stdio.h>
     
    
   

    

     

    
    

     

      void main()
     
    
   

    

     

    
    

     

      {
     
    
   

    

     

    
    

     
    int x,y,z;
     
    
   

    

     

    
    

     
    clrscr();
     
    
   

    

     

    
    

     
    puts("****************************************************************");
     
    
   

    

     

    
    

     
    puts("* This program is to solve Problem of Bridegroom and Bride. *");
     
    
   

    

     

    
    

     
    puts("* The Problem is as follows: Someone goes to 3 couples lovers' *");
     
    
   

    

     

    
    

     
    puts("* wedding. The bridegrooms are A,B,C and the brides are X,Y,Z. *");
     
    
   

    

     

    
    

     
    puts("* He wants to know who marries who and asks them. A says he *");
     
    
   

    

     

    
    

     
    puts("* will marry to X, X says her fiance is C, C says he will marry*");
     
    
   

    

     

    
    

     
    puts("* to Z. The man knows that they are all kidding. What they said*");
     
    
   

    

     

    
    

     
    puts("* is not true. So try to find who will marry to who? *");
     
    
   

    

     

    
    

     
    puts("****************************************************************");
     
    
   

    

     

    
    

     
    puts(" >> The solutions are:");
     
    
   

    

     

    
    

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

    

     

    
    

     
    for(x=1;x<=3;x++)          /*穷举x的全部可能配偶*/
     
    
   

    

     

    
    

     
        for(y=1;y<=3;y++)      /*穷举y的全部可能配偶*/
     
    
   

    

     

    
    

     
            for(z=1;z<=3;z++)    /*穷举z的全部可能配偶*/
     
    
   

    

     

    
    

     
                if(x!=1&&x!=3&&z!=3&&x!=y&&x!=z&&y!=z)  /*判断配偶是否满足题意*/
     
    
   

    

     

    
    

     

                      {
     
    
   

    

     

    
    

     
		    printf(" X will marry to %c.\n",'A'+x-1);    /*打印判断结果*/
     
    
   

    

     

    
    

     
		    printf(" Y will marry to %c.\n",'A'+y-1);
     
    
   

    

     

    
    

     
		    printf(" Z will marry to %c.\n",'A'+z-1);
     
    
   

    

     

    
    

     

                      }
     
    
   

    

     

    
    

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

    

     

    
    

     
    printf(" Press any key to quit...");
     
    
   

    

     

    
    

     
    getch();
     
    
   

    

     

    
    

     

      }
     
    
  
 

文章来源: blog.csdn.net，作者：程序员编程指南，版权归原作者所有，如需转载，请联系作者。

原文链接：blog.csdn.net/weixin_41055260/article/details/124536581