Jobdu-1336-液晶屏裁剪 | Acm之家


九度-1336-液晶屏裁剪 | Acm之家
苏州某液晶厂一直生产a * b大小规格的液晶屏幕,由于该厂的加工工艺限制,液晶屏的边长都为整数。最近由于市场需求发生变化,x : y比例的显示器比较流行,该厂为了适应市场需求,需要改造原生产线,并将库存的a * b大小的液晶屏进行裁剪,使之适应市场的需求。该厂同时希望,将a * b大小的液晶屏裁剪之后,不仅符合x : y的比例要求,同时在转换为新比例液晶屏的过程中,保证因切割而丢弃的液晶边角料最少。
你的任务就是根据原液晶屏大小和新的屏幕的比例,判断是否能将库存的液晶屏转换为符合要求的液晶屏大小,并求出新的液晶屏的大小a' b';若不能加工成新的,则输出0 0。
注:新的液晶屏边长和旧的液晶屏边长都为整数,并且长和宽不能交换。
输入:
测试数据包括多个,每个测试数据一行。
每行四个整数 a b x y
1) a、b表示原库存液晶屏的长和宽
2) x、y表示新的液晶屏的长宽比例x : y
其中,1 <= a, b, x, y <= 10^9。
输出:
对应每个测试案例,输出一行,每行包含两个整数,由空格隔开:
1)       若答案存在,则输出新的液晶屏的长宽 a' b'
2)       若不存在,则输出0 0
02int com_div(int a,int b)//辗转相除法
03{
04        int r;
05        while(b!=0)
06        {
07                r=a%b;
08                a=b;
09                b=r;
10        }
11        return a;
12}
13int main()
14{
15        int a,b,x,y,a1,b1,t,i;
16        while(scanf("%d%d%d%d",&a,&b,&x,&y)!=EOF)
17        {
18                t=com_div(x,y);
19                x=x/t;
20                y=y/t;
21                if(a<x||b<y)
22                {
23                        printf("0 0\n");
24                        continue;
25                }
26                if(x>y)
27                {
28                        for(a1=a/x*x;a1/x*y>b;a1-=x);
29                        b1=a1/x*y;
30                }
31                else
32                {
33                        for(b1=b/y*y;b1/y*x>a;b1-=y);
34                        a1=b1/y*x;
35                }
36                printf("%d %d\n",a1,b1);
37        }
38        return 0;
39}
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