链接: https://codeforces.com/contest/1176/problem/A 题意: You are given an integer n. You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times: Replace n with n2 if n is divisible by 2; Replace n with…

A. Divide it! 题目链接:http://codeforces.com/contest/1176/problem/A 题目 You are given an integer n You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times: Replace n with n2 if n is divisible by 2;Rep…

链接: https://codeforces.com/contest/1176/problem/B 题意: You are given an array a consisting of n integers a1,a2,-,an. In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter whe…

链接: https://codeforces.com/contest/1176/problem/C 题意: You are given an array a consisting of n integers. Each ai is one of the six following numbers: 4,8,15,16,23,42. Your task is to remove the minimum number of elements to make this array good. An a…

B. Merge it! 题目链接:http://codeforces.com/contest/1176/problem/B 题目 You are given an array a consisting of n integers a1,a2,…,an In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not…

题目地址:http://codeforces.com/contest/1176/problem/F 思路:其实就是一个01背包问题,只是添加了回合和每回合的01限制,和每当已用牌数到了10的倍数,那张卡会触发double攻击. 因为卡使用的多少会触发double效果,所以我们要记录攻击的同时记录卡的使用次数,可以由01背包dp[N][N]改变, 第一个维度是当前是第几个回合,第二个维度是记录卡在用了n张的情况下造成的攻击力,但是dp[N][N](N <= 2e5), 占用内存太大显然不行.于是想…

传送门 D. Divide by three, multiply by two •题意 给你一个数 x,有以下两种操作,x 可以任选其中一种操作得到数 y 1.如果x可以被3整除,y=x/3 2.y=x*2 y 再执行上述两种操作的一种得到数 z: 接着对 z 得到...... 这样依次执行了 n-1 次会得到 n 个数: 现在给你这 n 个数,让你按照上述规则给这 n 个数排序,使得其满足 a1=x , a2=y , a3=z , ........ •思路 对于任意一个数 p,能通过两类操作得…

传送门 A. Divide it! •题意 给定一个数n, 每次可以进行下列一种操作 1.如果n可以被2整除,用n/2代替n 2.如果n可以被3整除,用2n/3代替n 3.如果n可以被5整除,用4n/5代替n 如果可以经过上述操作使得 n 变为 1,输出最小操作次数,反之,输出-1: •思路 n/2 < 2n/3 < 4n/5 要想操作次数最少,优先操作 1 > 2 > 3: •代码 #include<bits/stdc++.h> using namespace std…

D. Recover it! Authors guessed an array aa consisting of nn integers; each integer is not less than 22 and not greater than 2⋅1052⋅105. You don't know the array aa, but you know the array bb which is formed from it with the following sequence of oper…

题意:给你一串只含\(4,8,15,16,23,42\)的序列,如果它满足长度是\(6\)的倍数并且有\(\frac {k}{6}\)个子序列是\([4,8,15,16,23,42]\),则定义它是好的,问最少删除多少元素使得序列是好的. 题解:我们开个桶(要离散化)记录这些数字出现的次数,然后线性遍历,当遇到\(4\),我们就直接让它++,否则判断它的前一个数的次数是否大于\(0\),如果是,那么它的前一个数的次数--,它自己次数++,如果前一个数出现的次数为\(0\),那么当前这个数就不能构…

题意:给你一个长度为\(n\)的序列\(a\).对它重新排列,使得\(a_{i+1}=a_{i}/3\)或\(a_{i+1}=2*a_{i}\).输出重新排列后的序列. 题解:经典DFS,遍历这个序列,对每个数dfs,每次dfs使\(len+1\),当\(len=n\)时,说明这个序列已经满足条件了,输出并且标记一下,防止还有另外的情况导致多输出. 代码: #include <iostream> #include <cstdio> #include <cstring>…

今天老师(orz sansirowaltz)让我们做了很久之前的一场Codeforces Round #257 (Div. 1),这里给出A~C的题解,对应DIV2的C~E. A.Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular cho…

 Codeforces Round #665 (Div. 2)  A. Distance and Axis 如果\(B\)在\(O\)左边,那么只能是定值\(OA\) 如果\(B\)在\(OA\)中间,那么必然小于等于\(OA\)且奇偶性和\(OA\)相同 \(B\)在\(A\)右边的情况显然不如\(B\)和\(A\)重合 所以分\(k\le n\)和\(k>n\)分类讨论即可 view code #pragma GCC optimize("O3") #pragma GCC op…

Codeforces Round #748 (Div. 3) A. Elections 思路分析: 令当前值比最大值大即可,如果最大值是它自己,就输出\(0\) 代码 #include <bits/stdc++.h> using namespace std; pair<int, int> a[3]; int ans[3]; int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); int t; cin >&g…

Codeforces Round #665 (Div. 2) 题解 写得有点晚了,估计都官方题解看完切掉了,没人看我的了qaq. 目录 Codeforces Round #665 (Div. 2) 题解 A - Distance and Axis B - Ternary Sequence C - Mere Array D - Maximum Distributed Tree E - Divide Square F - Reverse and Swap A - Distance and Axis…

& -- 位运算之一,有0则0 原题链接 Problem - 1514B - Codeforces 题目 Example input 2 2 2 100000 20 output 4 226732710 题意 t个测试样例,在每个样例中 数组有n个数,数字范围[ 0, 2k - 1] 使得数组每个数&后,结果=0,并且这n个数的和要尽量大 输出有多少个这样的数组 解析 数组每个数&后,结果=0  -->每一位至少一个0 数要尽量大  -->只要这一位可以 != 0, 就…

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…