[codeforces 260]B. Ancient Prophesy 试题描述 A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-". We'll say that some date is mentioned in the P…

260B - Ancient Prophesy 思路:字符串处理,把符合条件的答案放进map里,用string类中的substr()函数会简单一些,map中的值可以边加边记录答案,可以省略迭代器访问部分. 代码: #include<bits/stdc++.h> using namespace std; ; map<string,int>mp; string s; ]={,,,,,,,,,,,}; bool isOK(string s) { ]!=]!='-')return fals…

B. Ancient Berland Hieroglyphs 题目连接: http://codeforces.com/problemset/problem/164/B Descriptionww.co Polycarpus enjoys studying Berland hieroglyphs. Once Polycarp got hold of two ancient Berland pictures, on each of which was drawn a circle of hierog…

题目链接:http://codeforces.com/contest/456/problem/C 解题报告:给出一个序列,然后选择其中的一个数 k 删除,删除的同时要把k - 1和k + 1也删除掉,同时总分数里面加上一个k,求最大的分数可以是多少? dp题,递推公式是  dp[i] = max(dp[i-2]+num[i] * i,dp[i-1]);  ,注意要用long long ,一开始没用WA了两发. #include<cstdio> #include<cstring> #…

题目链接:http://codeforces.com/contest/456/problem/B 解题报告:输入一个n,让你判断(1n + 2n + 3n + 4n) mod 5的结果是多少?注意n的范围很大很大 n (0 ≤ n ≤ 10105). 只要判断是否能被整除4就可以了,如果n能被4整除,则结果是4,如果不能,则结果是0 但n很长,不能直接mod,但只要判断最低的两位能不能被4整除就可以了,如果n只有一位就判断最低的一位. #include<cstdio> #include<…

题目链接:http://codeforces.com/contest/456/problem/A 解题报告:有n种电脑,给出每台电脑的价格和质量,要你判断出有没有一种电脑的价格小于另一种电脑但质量却大于另一台电脑的情况. 把输入排个序就可以了,但是我比赛的时候排序只是按照价格排序了,锁定代码之后发现了这个错误,但已经改不了了,但是最后居然AC了,说明CF的数据也是有问题的. #include<cstdio> #include<cstring> #include<iostrea…

题目地址:http://codeforces.com/contest/456/problem/C 脑残了. .DP仅仅DP到了n. . 应该DP到10w+的. . 代码例如以下: #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdlib.h> #include <math.h> #include <cty…

C. Ancient Berland Circus time limit per test 2 seconds memory limit per test 64 megabytes input standard input output standard output Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.…

Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different. In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary fro…

C题就是个dp,把原数据排序去重之后得到新序列,设dp[i]表示在前i个数中取得最大分数,那么: if(a[i] != a[i-1]+1)   dp[i] = cnt[a[i]]*a[i] + dp[i-1]; else      dp[i] = dp[i] = max(dp[i-1],a[i]*cnt[a[i]] + dp[i-2]),    dp[i-1]表示不取a[i], a[i]*cnt[a[i]] + dp[i-2]表示取a[i]. cnt[a[i]]是a[i]出现的次数. #incl…

打表发现规律,对4取模为0的结果为4,否则为0,因此只需要判断输入的数据是不是被4整出即可,数据最大可能是100000位的整数,判断能否被4整出不能直接去判断,只需要判断最后两位(如果有)或一位能否被4整出即可. #include<map> #include<cmath> #include<queue> #include<cstdio> #include<string> #include<vector> #include<cst…

A:水的问题.排序结构.看看是否相同两个数组序列. B:他们写出来1,2,3,4,的n钍对5余.你会发现和5环节. 假设%4 = 0,输出4,否则输出0. 写一个大数取余就过了. B. Fedya and Maths time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Fedya studies in a gymnasium. Fe…

A. Adding Digits 枚举. B. Ancient Prophesy 字符串处理. C. Balls and Boxes 枚举起始位置\(i\),显然\(a_i \le a_j, 1 \le j \le n\), 同时\(i+1\)到\(x\)之间数需要大于\(a_i\),那么只要处理下区间最值即可. D. Black and White Tree 当\(s_i\)是最小值时,点\(i\)是个叶子节点,那么在不同颜色中的集合中找另一点连接即可. E. Dividing Kingdom…

A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-". We'll say that some date is mentioned in the Prophesy if there is a substring in the Pr…

D. New Year and Ancient Prophecy 题目连接: http://www.codeforces.com/contest/611/problem/C Description Limak is a little polar bear. In the snow he found a scroll with the ancient prophecy. Limak doesn't know any ancient languages and thus is unable to u…

C. Ancient Berland Circus 题目连接: http://www.codeforces.com/contest/1/problem/C Description Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different. In Ancient Berland arenas in circuses were s…

题目传送门 /* 题意:选择a[k]然后a[k]-1和a[k]+1的全部删除,得到点数a[k],问最大点数 DP:状态转移方程:dp[i] = max (dp[i-1], dp[i-2] + (ll) i * cnt[i]); 只和x-1,x-2有关,和顺序无关,x-1不取,x-2取那么累加相同的值,ans = dp[mx] */ #include <cstdio> #include <algorithm> #include <cstring> #include <…

题目传送门 /* DP:从1到最大值,dp[i][1/0] 选或不选,递推更新最大值 */ #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> using namespace std; typedef long long ll; ; const int INF = 0x3f3f3f3f; ll dp[MAXN][]; ll cnt[MAXN]; ll work(…

http://codeforces.com/contest/456/problem/A A. Laptops time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One day Dima and Alex had an argument about the price and quality of laptops. Dima thi…

D. Serega and Fun Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/455/problem/D Description Serega loves fun. However, everyone has fun in the unique manner. Serega has fun by solving query problems. One day Fedor came up w…

C. Civilization Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/455/problem/C Description Andrew plays a game called "Civilization". Dima helps him. The game has n cities and m bidirectional roads. The cities are numbe…

A. Boredom Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/455/problem/A Description Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and d…

题目链接:http://codeforces.com/problemset/problem/455/A 给你n个数,要是其中取一个大小为x的数,那x+1和x-1都不能取了,问你最后取完最大的和是多少. 简单dp,dp[i]表示取i时zui最大和为多少,方程为dp[i] = max(dp[i - 1] , dp[i - 2] + cont[i]*i). #include <bits/stdc++.h> using namespace std; typedef __int64 LL; ; LL a…

Fafa and Ancient Mathematics 转换成树上问题dp一下. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ…

1C - Ancient Berland Circus 思路: 求出三角形外接圆: 然后找出三角形三条边在小数意义下的最大公约数; 然后n=pi*2/fgcd; 求出面积即可: 代码: #include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define INF (1e9…

题目链接  Codeforces Round #465 (Div. 2) Problem E 题意  给定一个表达式,然后用$P$个加号和$M$个减号填充所有的问号(保证问号个数等于$P + M$) 求可以形成的表达式的最大值. 先把表达式转成一棵树,然后在树上DP. 题目保证了$min(P, M) <= 100$, 为了提高效率,我们选择用少的运算符号作为DP的第二维. 对$P$和$M$的大小关系进行分类讨论. 当$P < M$时, 设$f[i][j]$表示$i$代表的子树里面填$j$个加号…

题目链接:http://codeforces.com/problemset/problem/455/A A. Boredom time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alex doesn't like boredom. That's why whenever he gets bored, he comes up with…

time limit per test2.5 seconds memory limit per test512 megabytes inputstandard input outputstandard output Limak is a little polar bear. In the snow he found a scroll with the ancient prophecy. Limak doesn't know any ancient languages and thus is un…

题目链接:https://codeforces.com/contest/1362/problem/A 题意 有一个正整数 $a$,可选择的操作如下: $a \times 2$ $a \times 4$ $a \times 8$ $a\ /\ 2$,如果 $2$ 整除 $a$ $a\ /\ 4$,如果 $4$ 整除 $a$ $a\ /\ 8$,如果 $8$ 整除 $a$ 问能否由 $a$ 得到正整数 $b$,以及最少的操作次数. 题解 贪心模拟即可. 代码 #include <bits/stdc+…

题目链接:A.Johnny and Ancient Computer 题意: 给你两个数a,b.问你可不可以通过左移位运算或者右移位运算使得它们两个相等.可以的话输出操作次数,不可以输出-1 一次操作可以最多左移3次或者右移3次 题解: 首先找寻一下这两个数的二进制形式下最右边那个1在什么位置.然后看一下它们的差距是多少(设为x) 那么就让a,b中小的那个数左移x位.之后判断一下它们两个相等不相等就可以了 代码: 1 #include<stdio.h> 2 #include<algori…