链接: https://codeforces.com/contest/1265/problem/B 题意: You are given a permutation p=[p1,p2,-,pn] of integers from 1 to n. Let's call the number m (1≤m≤n) beautiful, if there exists two indices l,r (1≤l≤r≤n), such that the numbers [pl,pl+1,-,pr] is a…

题目链接:https://codeforces.com/contest/1265/problem/B 题意 给出大小为 $n$ 的一个排列,问对于每个 $i(1 \le i \le n)$,原排列中是否有一个大小为 $i$ 的连续子排列. 题解 从小到大构造排列,记录当前排列中数的最小下标和最大下标,若最小下标和最大下标的间距刚好为排列的长度,则说明大小为 $i$ 的排列是连续的. 代码一 #include <bits/stdc++.h> using namespace std; void s…

C. Beautiful Numbers 题目连接: http://www.codeforces.com/contest/300/problem/C Description Vitaly is a very weird man. He's got two favorite digits a and b. Vitaly calls a positive integer good, if the decimal representation of this integer only contains…

链接: https://codeforces.com/contest/1265/problem/D 题意: An integer sequence is called beautiful if the difference between any two consecutive numbers is equal to 1. More formally, a sequence s1,s2,-,sn is beautiful if |si−si+1|=1 for all 1≤i≤n−1. Trans…

链接: https://codeforces.com/contest/1265/problem/C 题意: So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved pi problems. Since…

链接: https://codeforces.com/contest/1265/problem/E 题意: Creatnx has n mirrors, numbered from 1 to n. Every day, Creatnx asks exactly one mirror "Am I beautiful?". The i-th mirror will tell Creatnx that he is beautiful with probability pi100 for al…

链接: https://codeforces.com/contest/1265/problem/A 题意: A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa&…

题目链接:https://codeforces.com/contest/1265/problem/E 题目大意: 有 \(n\) 个步骤,第 \(i\) 个步骤成功的概率是 \(P_i\) ,每一步只有成功了才会进入下一步,失败了会从第 \(1\) 步重新开始测.请问成功的期望步数是多少? 解题思路: 设期望步数是 \(S\) ,则有公式如下: 实现代码如下: #include <bits/stdc++.h> using namespace std; typedef long long ll;…

题目链接:https://codeforces.com/contest/1265/problem/A 题意 给出一个由 a, b, c, ? 组成的字符串,将 ? 替换为 a, b, c 中的一个字母,问能否字符串中所有相邻字母都不同. 题解 除非一开始字符串就不合法,否则一定可以构造出合法的字符串,因为共有三个字母可选,而替换时最多需要判断前后两个位置. 代码 #include <bits/stdc++.h> using namespace std; void solve() { strin…

题目链接:https://codeforces.com/contest/1265/problem/C 题意 从大到小给出 $n$ 只队伍的过题数,要颁发 $g$ 枚金牌,$s$ 枚银牌,$b$ 枚铜牌,要求: $g < s, g < b$ $g + s + d \le \lfloor \frac{n}{2} \rfloor$ 金牌队伍的过题数大于银牌,银牌队伍的过题数大于铜牌 输出颁发奖牌数最多的一种方案. 题解 根据第三个要求,奖牌一定是按过题数的大小颁发的. 金牌只颁发给过题最多数的队伍,…

题意 给出排成一列的 \(n\) 个格子,你要从 \(1\) 号格子走到 \(n\) 号格子之后(相当于 \(n+1\) 号格子),一旦你走到 \(i+1\) 号格子,游戏结束. 当你在 \(i\) 号格子时,你有 \(p_i\) 的概率走到 \(i+1\) 号格子,否则你会返回最近的一个 checkpoint (存档点),最近的存档点是指 \(max\{x|x\leq i \; and \; ischeckpoint(x)\}\) .对于每个 \(query x\) ,会把 \(x\) 号格子…

题目传送门 /* 构造:结构体排个序,写的有些啰嗦,主要想用用流,少些了判断条件WA好几次:( */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <vector> #include <map> #include <iostream> #include <string> using name…

A. Beautiful String (暴力) 题目链接 题目大意: 给定一个字符串,只有 \(?a\ b\ c\ ?\) ,问是否存在一种将所有的 \(?\) 替换成 \(a\ b\ c\) ,使得任意相邻的字符不同的方法. 大致思路: 其实可以发现问号都可以通过枚举使得其合法,若出现不合法必然除去问好已经不合法,暴力枚举即可. 代码: 点击展开代码 #include<bits/stdc++.h> using namespace std; const int N=1e5+10; char…

链接:http://codeforces.com/contest/1265 A. Beautiful String A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "a…

B. Beautiful Paintings 题目连接: http://www.codeforces.com/contest/651/problem/B Description There are n pictures delivered for the new exhibition. The i-th painting has beauty ai. We know that a visitor becomes happy every time he passes from a painting…

E. Binary Numbers AND Sum 题目链接:https://codeforces.com/contest/1066/problem/E 题意: 给出两个用二进制表示的数,然后将第二个二进制不断地往右边移一位,每次答案加上这两个的交集,求最后的答案. 题解: 考虑第二个二进制每一位对答案的贡献就行了,然后对第一个二进制算前缀和就ok了. 代码如下: #include <bits/stdc++.h> using namespace std; typedef long long l…

链接: https://codeforces.com/contest/1182/problem/C 题意: You are given n words, each of which consists of lowercase alphabet letters. Each word contains at least one vowel. You are going to choose some of the given words and make as many beautiful lyric…

题目链接:http://www.codeforces.com/problemset/problem/271/A题意:给你一个四位数,求比这个数大的最小的满足四个位的数字不同的四位数.C++代码: #include <iostream> #include <algorithm> using namespace std; bool chk(int x) { ]; ; i < ; i ++) { a[i] = x % ; x /= ; } sort(a, a + ); ; i &l…

A. Beautiful Year time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after t…

B. Beautiful Paintings time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n pictures delivered for the new exhibition. The i-th painting has beauty ai. We know that a visitor becomes…

// https://codeforces.com/contest/1265/problem/D /* 感觉像是遍历的思维构造题 有思路就很好做的 可以把该题想象成过山车或者山峰...... */ #include<iostream> #include<cstdio> using namespace std; int n; int cnt[5], last[5]; // last 是记录当前还有多少 0, 1, 2, 3 int ans[100005]; bool ok; int…

Beautiful Sequence Beautiful Mirrors Beautiful Bracket Sequence (easy version) Beautiful Sequence \[ Time Limit: 1000 ms\quad Memory Limit: 256 MB \] 首先我们可以考虑到 \(0\) 只能 和 \(1\) 放在一起.\(3\) 只能和 \(2\) 放在一起,那么我们想办法先把 \(0\) 和 \(3\) 凑出来,最后就剩下 \(1\) 和 \(2\)…

A. Magic Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of…

题目链接:https://codeforces.com/contest/1417/problem/C 题意 给出一个大小为 $n$ 的数组 $a$,计算当 $k$ 从 $1$ 到 $n$ 取值时在所有 $k$ 长区间中出现的数的最小值. 题解 记录一个值两两间的最大距离,该距离的 $k$ 长区间及之后更长的区间都可能以该值为最小值. Tips 注意两端的处理. 代码一 #include <bits/stdc++.h> using namespace std; int main() { ios:…

Problem - D - Codeforces 题意 : 有 a 个0,b个1,c个2,d个3,构成一个序列,使得每两个数字之间的差值为1 题解: 就是以四种数字分别为起点,暴力模拟 #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef pair<ll,ll> pll; const int N=4e5+50; const ll inf=1e18; const ll mod=998244…

#include <iostream> #include <vector> #include <algorithm> #include <string> using namespace std; int main(){ long long n; cin >>n; while(n){ if(n%10 == 1) n/=10; else if(n%100 == 14 ) n/=100; else if(n%1000 == 144) n/=1000;…

A题 链接 思路分析: 因为只需要做到相邻的不相同,利用三个不同的字母是肯定可以实现的, 所以直接先将所有的问号进行替换,比如比前一个大1,如果与后面的冲突,则再加一 代码(写的很烂): #include <bits/stdc++.h> using namespace std; int check( string a) { for (int i = 0; i < a.length(); ++i) { if (i==0) { if (a[i]==a[i+1]) { return 0; }…

比赛传送门 感觉这场是最近以来做过的最顺手的一场,持续上分,开心w A了 前三题,然后第四题其实还有半个多小时,但怕身体撑不住,就先退了,其实第四题也很简单 自己认为的算法标签: ​ A.暴力模拟.字符串 ​ B.数学.构造.排序 ​ C.贪心.构造 ​ D.构造.遍历 :D A. Beautiful String // https://codeforces.com/contest/1265/problem/A /* 要求:不能有连续相同的字符,然后把'?'改为'a' || 'b' || 'c'…

思路:直接暴力判断就OK了 #include<bits/stdc++.h> using namespace std; #define int long long signed main(){ int _;cin>>_; while(_--){ string str; cin>>str; ){ ]!='?'){ cout<<str<<'\n'; }else{ printf("a\n"); } continue; } ; ;i&l…

构造,枚举起点,如果一个序列成立,那么将它reverse依然成立,所以两个方向(从小到大或从大到小)没有区别,选定一个方向进行探测,直到探测不到以后回头,只要所给数据能成立,那么能探测进去就能探测出来,否则就不能构造. #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; int sum; int main(){ ios::sync_with_stdio(false); cin.tie(NULL…