Description You are given a positive integer nn. Let S(x) be sum of digits in base 10 representation of xx, for example, S(123)=1+2+3=6, S(0)=0. Your task is to find two integers a,ba,b, such that 0≤a,b≤n, a+b=n and S(a)+S(b) is the largest possible…

Maximum Sum of Digits You are given a positive integer n. Let S(x)S(x) be sum of digits in base 10 representation of xx , for example, S(123)=1+2+3=6S(123)=1+2+3=6 , S(0)=0S(0)=0 . Your task is to find two integers a,ba,b , such that 0≤a,b≤n0≤a,b≤n ,…

(CodeForces - 5C)Longest Regular Bracket Sequence time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output This is yet another problem dealing with regular bracket sequences. We should remind you t…

B. Maximum Sum of Digits time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output You are given a positive integer nn. Let S(x)S(x) be sum of digits in base 10 representation of xx, for example, S(…

B - Sorted Adjacent Differences(CodeForces - 1339B) 题目链接 算法 思维+贪心 时间复杂度O(nlogn) 1.这道题的题意主要就是让你对一个数组进行一种特殊的排序,使得数组中相邻的两个数的差的绝对值成非递减趋势: 2.刚开始对这道题总是执拗于两个相等的数在不同位置,如何把它们放到前面这个问题,因为路走歪了,最终无果,没有思路.后来看了一些关于这道题的解题博客,豁然开朗. 3.使得数组中相邻的两个数的差的绝对值成非递减趋势,怎么想呢.单纯想怎么…

Given Length and Sum of Digits... 题目链接: http://acm.hust.edu.cn/vjudge/contest/121332#problem/F Description You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the largest of the numbers that have length m…

C. Given Length and Sum of Digits... time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the la…

题目链接:https://codeforces.com/problemset/problem/1197/D 题意: 给你一个序列,求一个子序列 a[l]~a[r] 使得该子序列的 sum(l,r)-k*(r-l+1)/m(向上取整)的值是在所有子序列中最大的,并输出最大值 思路: 法一:动态规划 dp[i][j] 表示序列到i截止,这一轮已经进行了j次取数(j = (len+m-1)%m) 那么dp[i][j]维护的就是起点为 s = i-j+1-m*t (t>=0)这个集合的最优,这样所有的…

题目链接:http://codeforces.com/problemset/problem/558/C 题意:给出n个数,让你通过下面两种操作,把它们转换为同一个数.求最少的操作数. 1.ai = ai*2 2.ai = ai/2,向下取整 思路: 可以除以二 或者 乘以二,就相当于位运算的右移和左移.用两个数组,vis 数组, cnt 数组.刚开始都初始化为0: vis[i] 表示 i 这个数可以由几个数转化而来,cnt[i] 表示题目给出的 n 个数全部转化为 i 需要的操作数. 首先遍历数…

转载请注明出处:http://blog.csdn.net/u012860063 Max Sum of Max-K-sub-sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5791    Accepted Submission(s): 2083 Problem Description Given a circle seq…