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<…

http://codeforces.com/contest/456/problem/B CF#260 div2 B Fedya and Maths Codeforces Round #260 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 gymnasi…

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. Fedya's maths hometask is to calculate the following expression: (1n + 2n + 3n + 4n) mod 5 f…

Description Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression: (1n + 2n + 3n + 4n) mod 5 for given value of n. Fedya managed to complete the task. Can you? Note that given number n can be extremely large (e…

[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…

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…

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/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…

The Maths Lecture 题意:求存在后缀Si mod k =0,的n位数的数目.(n <=1000,k<=100); 用f[i][j]代表 长为i位,模k等于j的数的个数. 可以用 f[i+1][(t*10i+j)%k]=∑f[i][j]+(j==0),(t*10i+j)%k!=0;动态规划 这样可以求出所有f[n][i] i>0 的值. 最后用9*10^(n-1)-∑f[n][i] 就可以得到 答案 #include <bits/stdc++.h> using…

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…

题目链接:http://www.codeforces.com/problemset/problem/339/A题意:重新组合加法字符串,使得按照1,2,3的顺序进行排列.C++代码: #include <iostream> #include <string> using namespace std; ]; string s, ans = ""; int main() { cin >> s; int len = s.length(); ; i <…

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…

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…

A. Laptops 题目意思: 给定n台电脑,第i台电脑的价格是ai ,质量是bi ,问是否存在一台电脑价格比某台电脑价格底,但质量确比某台电脑的质量高,即是否存在ai < aj 且 bi > bj ? 解题思路: 这题一定要看题目,a都是1~n的不同数,b也是1~n的不同数,此题只需要判断ai 是否等于bi ,如果ai != bi 的话,则输出“Happy Alex”,如果所有的ai  == bi 则输出“Poor Alex” 证明:先将a按照从小到大排序,当i<j时ai <…

Description Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression: (1n + 2n + 3n + 4n) mod 5 for given value of n. Fedya managed to complete the task. Can you? Note that given number n can be extremely large (e…

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 thinks that the more expensive a laptop is, the…

题目链接 A. Laptops time limit per test:1 secondmemory limit per test:256 megabytesinput:standard inputoutput:standard output One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, t…

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 thinks that the more expensive a laptop is, the…

考试考到自闭,每天被吊打. 还有几天可能就要AFO了呢... Luogu3602:Koishi Loves Segments 从左向右,每次删除右端点最大的即可. [HEOI2014]南园满地堆轻絮 答案一定是 \(\lceil \frac{max_{1\le i < j \le n}(a_i-a_j)}{2} \rceil\). 可以考虑一个二分答案 \(mid\),那么每个数 \(x\) 都是一个 \([x-mid,x+mid]\) 的范围. 当前面有一个 \(y\) 使得 \(y-mid>…

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(…

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…

题目链接: http://codeforces.com/contest/1152/problem/C 题意: 给出两个数$a$和$b$ 找一个$k(k\geq 0)$得到最小的$LCM(a+k,b+k)$ 如果有多个$k$,输出最小的$k$ 数据范围: $1 \le a, b \le 10^9$ 分析: 假设 $gcd\left (  a+k,b+k\right )= t$ 那么$(a+k)\%t=(b+k)\%t=0$ 化简得到$a\%t=b\%t$ $a-x\times t=b-y\time…

题目:http://codeforces.com/contest/1152/problem/C 题意:给你a,b, 你可以找任意一个k     算出a+k,b+k的最小公倍数,让最小公倍数尽量小,求出这个k 思路: 因为现在两个都是未知数,我们无法确定 我们根据gcd底层实现原理 gcd(a+k,b+k) = gcd(b-a,a+k) a=c*x; b=c*y; b-a=c*(y-x) 所以证明b-a的因子是a的因子也是b的因子 那么我们只要枚举出b-a的因子,然后再套入a+k中求得k,然后枚举…

传送门 D. The Maths Lecture time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr doesn't like Maths as he finds it really boring, so he usually sleeps in Maths lectures. But one day the teacher…

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…