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那精度怎么处理？

Posted by madongtest at 2006-08-20 02:19:15 on Problem 2976
In Reply To:我的方法（in），枚举+贪心。有点慢 Posted by:xiaolonghingis at 2006-08-19 15:33:35

> 因为答案c只可能在100到0之间，从大到小枚举c。假设答案为c，那么必然有                   
>      100*(a1+.....+ak)
>     -------------------- 〉=c ,两边同乘分母后得
>        b1+.....+bk
>     100*(a1+.....+ak)-c*(b1+.......+bk)>=0
> 
>     取100*ai-c*bi中最大的n-k个，看他们的和是否大于零即可。

Followed by:

这里都是整数的运算，不用处理什么精度吧 (0) kevinew 2006-08-29 01:18:59

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