ans=max(sum+abpre[i-j+1-1]+absuf[i+j+1],ans); } } cout << ans;return0; }
Codeforces 631E Product Sum 斜率优化 我们先把问题分成两部分, 一部分是把元素往前移, 另一部分是把元素往后移。对于一个 i 后的一个位置, 我们考虑前面哪个移到这里来最优。 我们设最优值为val, val = max(a[ j ] * (i - j) - (sum[ i ] - sum[ j ]) 我们能发现这个能转换成斜率优化的形...
Codeforces 631E Product Sum 斜率优化 我们先把问题分成两部分, 一部分是把元素往前移, 另一部分是把元素往后移。对于一个 i 后的一个位置, 我们考虑前面哪个移到这里来最优。 我们设最优值为val, val = max(a[ j ] * (i - j) - (sum[ i ] - sum[ j ]) 我们能发现这个能转换成斜率优化的形...
Samples 输入数据 1 4 4 3 2 5 Copy 输出数据 1 39 Copy 输入数据 2 5 1 1 2 7 1 Copy 输出数据 2 49 Copy Note In the first sample, one may pick the first element and place it before the third (before5). Thus, the answer will be3·1 + 2·2 + 4·3 + 5·4...
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Codeforces Round #344 (Div. 2) E. Product Sum 维护凸壳 题目连接: http://www.codeforces.com/contest/631/problem/E Description Blake is the boss of Kris, however, this doesn't spoil their friendship. They often gather at the bar to talk about intriguing problems about maximising some values...
codeforces 1269 E K Integers 2019-12-24 03:42 −# E. K Integers ## 题目连接:https://codeforces.com/contest/1269/problem/E ## 题意 给了一个排列p,你每次操作可以交换两个相邻的元素,现在问你最少操作多少次可以形成一个形如1,2,3,4..k 这样的子段 k从1~n ## 题解: ... ...
Given an integer number n, return the difference between the product of its digits and the sum of its digits. Example 1: Input: n = 234 Output: 15 Explanation: Product of digits = 2 * 3 * 4 = 24 Sum of digits = 2 + 3 + 4 = 9 Result = 24 - 9 = 15 ...
E. Product Sum time limit per test: 1 second memory limit per test: 256 megabytes input:standard input output:standard output E. Product Sum time limi
@codeforces - 631E@ Product Sum 目录 @desription@ @solution@ @accepted code@ @details@ @desription@ 给定一个序列 a,定义它的权值c=∑ni=1aic=∑i=1nai。 你可以做如下的操作恰好一次:选择一个数,然后将它移动到一个位置(可以是原位置,序列开头与结尾)。