085.百钱百鸡问题
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#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();
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