085.百钱百鸡问题

  

   

    

     

    
    

     

      #include<stdio.h>
     
    
   

    

     

    
    

     

      void main()
     
    
   

    

     

    
    

     

      {
     
    
   

    

     

    
    

     
    int x,y,z,j=0;
     
    
   

    

     

    
    

     
    clrscr();
     
    
   

    

     

    
    

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

    

     

    
    

     
    puts("* This program is to solve Problem of *");
     
    
   

    

     

    
    

     
    puts("* Hundred Yuan Hundred Fowls. *");
     
    
   

    

     

    
    

     
    puts("* Which is presented by Zhang Qiujiang, *");
     
    
   

    

     

    
    

     
    puts("* a Chinese ancient mathematician, in his work *");
     
    
   

    

     

    
    

     
    puts("* Bible of Calculation: 5 Yuan can buy 1 cock, *");
     
    
   

    

     

    
    

     
    puts("* 3 Yuan can buy 1 hen, 1 Yuan buy 3 chickens, *");
     
    
   

    

     

    
    

     
    puts("* now one has 100 Yuan to buy 100 fowls, the *");
     
    
   

    

     

    
    

     
    puts("* question is how many cocks, hens, chickens *");
     
    
   

    

     

    
    

     
    puts("* to buy? *");
     
    
   

    

     

    
    

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

    

     

    
    

     
    printf("\n The possible plans to buy 100 fowls with 100 Yuan are:\n\n");
     
    
   

    

     

    
    

     
    for(x=0;x<=20;x++)               /*外层循环控制鸡翁数*/
     
    
   

    

     

    
    

     
        for(y=0;y<=33;y++)           /*内层循环控制鸡母数y在0~33变化*/
     
    
   

    

     

    
    

     

              {
     
    
   

    

     

    
    

     

                  z=100-x-y;             /*内外层循环控制下，鸡雏数z的值受x,y的值的制约*/
     
    
   

    

     

    
    

     
            if(z%3==0&&5*x+3*y+z/3==100)
     
    
   

    

     

    
    

     
                                   /*验证取z值的合理性及得到一组解的合理性*/
     
    
   

    

     

    
    

     
                printf("%2d: cock=%2d hen=%2d chicken=%2d\n",++j,x,y,z);
     
    
   

    

     

    
    

     

      	}
     
    
   

    

     

    
    

     
    puts("\n 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/124518702