[HDOJ]1009. FatMouse' Trade
恩,简单的排序问题,以前的排序问题只是会牵扯一个数组,这次只不过是把两个数组结合起来排序而已。
为了以最小的F[i]得到最多的J[i],只需要按照F[i]/J[i]的值从大到小对两个数组重新排序下就可以了。然后再按照从大到小的数组去分配M pounds of cat food,直到food分配完或者所有的JavaBeans都被保护为止。AC代码如下:
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我没有什么雄心壮志,我只想给自己和关心自己的家人和朋友一个交代,仅此而已。