摘要: 题目大意:有n个物体,大小为$c_i$。把第i个到第j个放到一起,容器的长度为$x=j-i+\sum\limits_{k-i}^{j} c_k$,若长度为x,费用为$(x-L)^2$。费用最小. 题解: $$令:a_i=\sum\limits_{i=1}^{i} c_i$$ $$dp_i=min(d 阅读全文
posted @ 2018-06-12 16:31 Memory_of_winter 阅读(161) 评论(0) 推荐(0) 编辑