E - Superior Periodic Subarrays 好难的一题啊... 这个博客讲的很好,搬运一下. https://blog.csdn.net/thy_asdf/article/details/49406133 #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define pii pair<int,int> #de…

B. Once Again... Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/582/problem/B Description You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai -…

C. GCD Table Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/583/problem/C Description The GCD table G of size n × n for an array of positive integers a of length n is defined by formula Let us remind you that the greatest c…

http://codeforces.com/contest/583/problem/D 原题:You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the give…

D. Once Again... time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that …

题目链接:http://codeforces.com/contest/583/problem/C C. GCD Table time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The GCD table G of size n × n for an array of positive integers a of length n …

C. GCD Table The GCD table G of size n × n for an array of positive integers a of length n is defined by formula Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both…

A. GCD Table time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The GCD table G of size n × n for an array of positive integers a of length n is defined by formula Let us remind you that the…

A[i][j]表示在循环节下标i开头j结尾的最长不减子序列,这个序列的长度为p,另外一个长度为q的序列对应的矩阵为B[i][j], 将两序列合并,新的序列对应矩阵C[i][j] = max(A[i][k]+B[k][j]).非法的情况标记为-INF,用倍增加速. #include<bits/stdc++.h> using namespace std; const int INF = 0x3f3f3f3f; ; int n; typedef int MType; struct Matrix {…

对角线上的元素就是a[i],而且在所在行和列中最大, 首先可以确定的是最大的元素一定是a[i]之一,这让人想到到了排序. 经过排序后,每次选最大的数字,如果不是之前更大数字的gcd,那么只能是a[i]之一. div2路漫漫... #include<bits/stdc++.h> using namespace std; typedef int ll; ll a[]; *]; ll gcd(ll a,ll b) { return b?gcd(b,a%b):a; } map<int,int&g…

C. GCD Table time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The GCD table G of size n × n for an array of positive integers a of length n is defined by formula Let us remind you that the…

B. Robot's Task time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Robot Doc is located in the hall, with n computers stand in a line, numbered from left to right from 1 to n. Each computer co…

A. Asphalting Roads time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output City X consists of n vertical and n horizontal infinite roads, forming n × n intersections. Roads (both vertical and hori…

被进爷坑了,第二天的比赛改到了12点 水 A - Asphalting Roads /************************************************ * Author :Running_Time * Created Time :2015/10/3 星期六 21:53:09 * File Name :A.cpp ************************************************/ #include <cstdio> #include…

                                                                              D. Once Again... You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of…

B. Once Again... time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that …

Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…

Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…

直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…

 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…

Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/output 1 s, 256 MB  x2377 B Queue standard input/output 2 s, 256 MB  x1250 C Hacking Cypher standard input/output 1 s, 256 MB  x740 D Chocolate standard in…

Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his info…

Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Victor adores the sets theory. Let us remind you that a set is a group of…

A: 题目大意: 在一个multiset中要求支持3种操作: 1.增加一个数 2.删去一个数 3.给出一个01序列,问multiset中有多少这样的数,把它的十进制表示中的奇数改成1,偶数改成0后和给出的01序列相等(比较时如果长度不等各自用0补齐) 题解: 1.我的做法是用Trie数来存储,先将所有数用0补齐成长度为18位,然后就是Trie的操作了. 2.官方题解中更好的做法是,直接将每个数的十进制表示中的奇数改成1,偶数改成0,比如12345,然后把它看成二进制数10101,还原成十进制是2…

CF469 Codeforces Round #268 (Div. 2) http://codeforces.com/contest/469 开学了,时间少,水题就不写题解了,不水的题也不写这么详细了. A 水题 //#pragma comment(linker, "/STACK:102400000,102400000") #include<cstdio> #include<cmath> #include<iostream> #include<…

题目传送门 /* 贪心 + 模拟:首先,如果蜡烛的燃烧时间小于最少需要点燃的蜡烛数一定是-1(蜡烛是1秒点一支), num[g[i]]记录每个鬼访问时已点燃的蜡烛数,若不够,tmp为还需要的蜡烛数, 然后接下来的t秒需要的蜡烛都燃烧着,超过t秒,每减少一秒灭一支蜡烛,好!!! 详细解释:http://blog.csdn.net/kalilili/article/details/43412385 */ #include <cstdio> #include <algorithm> #i…

题目传送门 /* 题意:从前面找一个数字和末尾数字调换使得变成偶数且为最大 贪心:考虑两种情况:1. 有偶数且比末尾数字大(flag标记):2. 有偶数但都比末尾数字小(x位置标记) 仿照别人写的,再看自己的代码发现有清晰的思维是多重要 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include…

#include <iostream> #include <string> using namespace std; int main(){ int n; cin >> n; string str; cin >> str; , x = ; ; i < n ; ++ i){ if(str[i] == 'B') cnt+=(x << i); } cout<<cnt<<endl; }   Codeforces Round…

Codeforces Round #160 (Div. 1) A - Maxim and Discounts 题意 给你n个折扣,m个物品,每个折扣都可以使用无限次,每次你使用第i个折扣的时候,你必须买q[i]个东西,然后他会送你{0,1,2}个物品,但是送的物品必须比你买的最便宜的物品还便宜,问你最少花多少钱,买完m个物品 题解 显然我选择q[i]最小的去买就好了 代码 #include<bits/stdc++.h> using namespace std; const int maxn =…

Codeforces Round #383 (Div. 2) A. Arpa's hard exam and Mehrdad's naive cheat 题意 求1378^n mod 10 题解 直接快速幂 代码 #include<bits/stdc++.h> using namespace std; long long quickpow(long long m,long long n,long long k) { long long b = 1; while (n > 0) { if…