CF719C. Efim and Strange Grade[DP]】的更多相关文章

C. Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any po…
Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any posit…
C. Efim and Strange Grade 题目连接: http://codeforces.com/contest/719/problem/C Description Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any positive decimal fraction. First he got disappoin…
C. Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any po…
题目链接:http://codeforces.com/problemset/problem/719/C C. Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a…
codeforces 373 A - Efim and Strange Grade(算数模拟) 原题:Efim and Strange Grade 题意:给出一个n位的实型数,你可以选择t次在任意位进行四舍五入的进位,求最大结果. 解法:这道题一定不能忽略数位计算时本身带来的进位,如果我们要改变这个数的大小,一定是在最先的那个出现5以上的数字进行四舍五入,之后的t次允许我们多次四舍五入,如果自然进位则不消耗t. 最后一点,整数位的进位也是需要考虑的 #include <cstdio> #inc…
Codeforces 718A Efim and Strange Grade 程序分析 jerry的程序 using namespace std; typedef long long ll; string buf; int i; void up(int at) { at--; if (at < 0) { buf = '1' + buf; i++; return; } if (buf[at] == '.') at--; buf[at]++; if (buf[at] == '9'+1) { buf[…
time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any positive decimal fraction. F…
题意:给定一个浮点数,让你在时间 t 内,变成一个最大的数,操作只有把某个小数位进行四舍五入,每秒可进行一次. 析:贪心策略就是从小数点开始找第一个大于等于5的,然后进行四舍五入,完成后再看看是不是还可以,一循环下去,直到整数位,或者没时间了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <…
[题解]Making The Grade(DP+结论) VJ:Making the Grade HNOI-D2-T3 原题,禁赛三年. 或许是我做过的最简单的DP题了吧(一遍过是什么东西) 之前做过关于绝对值的题目,这种要求绝对值最小的题目,有一个很普遍的结论,最优解的集合中,一定有一个满足所有元素一定是所给定的元素中的元素,具体证明或许就是把括号拆开或者反证法吧. 然后就是这种看起来是\(O(n^3)\)的DP可以通过巧妙的实现降到\(O(n^2)\),当然你暴力使用数据结构变成\(O(n^2…
Making the Grade Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6445   Accepted: 2994 Description A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up…
Description A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up or down a single slope, but they are not fond of an alternating succession of hills and valleys. FJ…
题目传送门 Making the Grade Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9090   Accepted: 4253 Description A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbi…
题意:给你一组数,问你最少删去多少数,使得剩下的数,每个数都能整除数组中其它某个数或被数组中其它某个数整除. 题解:我们直接枚举所有因子,\(dp[i]\)表示\(i\)在数组中所含的最大因子数(当我们枚举到\(i\)时),然后用\(dp[i]\)更新以\(i\)作为因子的更大的数,注意,更新的时候\(dp[j]=max(dp[i],dp[j])\),而不是\(dp[j]+=dp[i]\),因为这样会把之前的因子重复计算. 代码: #include <bits/stdc++.h> #defin…
A. Vitya in the Countryside time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every gran…
Codeforces Round #373 (Div. 1) A. Efim and Strange Grade 题意 给一个长为\(n(n \le 2 \times 10^5)\)的小数,每次可以选择某位小数进行四舍五入,最多做\(t(t \le 10^9)\)次. 求最大的数. 思路 每次必然找小数部分能进行四舍五入的最高位,并且四舍五入的位置是递增的. 注意:99.5这样的数据,最高位会进位. 代码 A. Efim and Strange Grade C. Sasha and Array…
A. Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any po…
用了两场比赛上Div 1感觉自己好腊鸡的说...以下是这两场比赛的部分题解(不得不说有个黄学长来抱大腿还是非常爽的) Round #372 : Div 2 A:Crazy Computer 题意:给定N个输入和一个时间长度M,每次输入屏幕上增加一个字符,若两个输入间隔大于M则屏幕上的字符会被清空,问结束时屏幕上还有多少个字符 直接模拟没有什么好说的 代码: #include<cstdio> #include<iostream> #include<cstring> #in…
A - Complete the Word(暴力) Description ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once.…
A. Vitya in the Countryside time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every gran…
A. Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any po…
C. Efim and Strange Grade Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any positive decimal fraction. First he got disappointed, as he expected a way more pleasant result. Then, he devel…
收录已发布的题解 按发布时间排序. 部分可能与我的其他文章有重复捏 qwq . AtCoder for Chinese: Link ZHOJ: Link 洛谷 \(1\sim 5\) : [题解]CF45I TCMCF+++ [题解]CF1013B And [题解]CF991C Candies [题解]CF356A Knight Tournament [题解]CF1715A Crossmarket \(6\sim 10\) : [题解]CF1215C Swap Letters [题解]CF172…
题目大意:每次给出两个碱基序列(包含ATGC的两个字符串),其中每一个碱基与另一串中碱基如果配对或者与空串对应会有一个分数(可能为负),找出一种方式使得两个序列配对的分数最大 思路:字符串动态规划的经典题,很容易想到状态dp[i][j],指第一个长度为i的串和第二个长度为j的串配对的最大分数.显然,这个状态可以由dp[i][j-1],dp[i-1][j],dp[i-1][j-1]三个子问题得到,即第一串最后一个字符对应空格.第二串最后一个字符对应空格和第一串第二串最后一个字符配对所得到的分数这三…
最优的做法最后路面的高度一定是原来某一路面的高度. dp(x, t) = min{ dp(x - 1, k) } + | H[x] - h(t) | ( 1 <= k <= t ) 表示前 i 个路面单调不递减, 第 x 个路面修整为原来的第 t 高的高度. 时间复杂度O( n³ ). 令g(x, t) = min{ dp(x, k) } (1 <= k <= t), 则转移O(1), g() 只需在dp过程中O(1)递推即可, 总时间复杂度为O( n² ) 然后单调不递增也跑一遍…
Description A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up or down a single slope, but they are not fond of an alternating succession of hills and valleys. FJ…
Description A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up or down a single slope, but they are not fond of an alternating succession of hills and valleys. FJ…
Making the Grade Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 1 Problem Description A straight dirt road connects two fields on FJ's farm, but it changes elevatio…
Making the Grade Time Limit: 1000MS   Memory Limit: 65536K Total Submissions:10187   Accepted: 4724 Description A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up…
    正常的没想到的DP和玄学贪心. 题目描述 A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up or down a single slope, but they are not fond of an alternating succession of hills and…