Codeforces Round #277 (Div. 2) 题解_html/css_WEB-ITnose
Codeforces Round #277 (Div. 2)
A. Calculating Function
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
For a positive integer n let's define a function f:
f(n)?=??-?1?+?2?-?3?+?..?+?(?-?1)nn
Your task is to calculate f(n) for a given integer n.
Input
The single line contains the positive integer n (1?≤?n?≤?1015).
Output
Print f(n) in a single line.
Sample test(s)
input
output
input
output
-3
Note
f(4)?=??-?1?+?2?-?3?+?4?=?2
f(5)?=??-?1?+?2?-?3?+?4?-?5?=??-?3
简单公式:
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long int LL;LL n;int main(){ cin>>n; if(n%2==0) { LL t=n/2; cout <br> <br> <p></p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> B. OR in Matrix </p> <p class="sycode"> </p> <p class="sycode"> time limit per test </p> 1 second <p class="sycode"> </p> <p class="sycode"> memory limit per test </p> 256 megabytes <p class="sycode"> </p> <p class="sycode"> input </p> standard input <p class="sycode"> </p> <p class="sycode"> output </p> standard output <p class="sycode"> </p> <p> Let's define logical OR as an operation on two logical values (i. e. values that belong to the set {0,?1}) that is equal to 1 if either or both of the logical values is set to 1, otherwise it is 0. We can define logical OR of three or more logical values in the same manner:</p> <p> where is equal to 1 if some ai?=?1, otherwise it is equal to 0.</p> <p> Nam has a matrix A consisting of m rows and n columns. The rows are numbered from 1 to m, columns are numbered from 1 to n. Element at row i(1?≤?i?≤?m) and column j (1?≤?j?≤?n) is denoted as Aij. All elements of A are either 0 or 1. From matrix A, Nam creates another matrix B of the same size using formula:</p> <p> .</p> <p> (Bij is OR of all elements in row i and column j of matrix A)</p> <p> Nam gives you matrix B and challenges you to guess matrix A. Although Nam is smart, he could probably make a mistake while calculating matrix B, since size of A can be large.</p> <p class="sycode"> </p> <p class="sycode"> Input </p> <p> The first line contains two integer m and n (1?≤?m,?n?≤?100), number of rows and number of columns of matrices respectively.</p> <p> The next m lines each contain n integers separated by spaces describing rows of matrix B (each element of B is either 0 or 1).</p> <p class="sycode"> </p> <p class="sycode"> Output </p> <p> In the first line, print "NO" if Nam has made a mistake when calculating B, otherwise print "YES". If the first line is "YES", then also print m rows consisting of n integers representing matrix A that can produce given matrix B. If there are several solutions print any one.</p> <p class="sycode"> </p> <p class="sycode"> Sample test(s) </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> input </p> <pre style="代码" class="precsshei">2 21 00 0
output
NO
input
2 31 1 11 1 1
output
YES1 1 11 1 1
input
2 30 1 01 1 1
output
YES0 0 00 1 0
水题:将A里所有可能是1的点加上就可以了
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n,m;int B[220][220];int A[220][220];bool vis[220][220];bool check(int x,int y){ bool flag=true; for(int i=0;i<m if flag="false;" for i="0;i<n&&flag;i++)" return cl x y vis main cin>>n>>m; for(int i=0;i<n for j="0;j<m;j++)" cin>>B[i][j]; for(int i=0;i<n for j="0;j<m;j++)" if a cl bool flag="true;" i="0;i<n&&flag;i++)" puts cout else return> <br> <br> <p class="sycode"> <br> </p> <p class="sycode"> C. Palindrome Transformation </p> <p class="sycode"> </p> <p class="sycode"> time limit per test </p> 1 second <p class="sycode"> </p> <p class="sycode"> memory limit per test </p> 256 megabytes <p class="sycode"> </p> <p class="sycode"> input </p> standard input <p class="sycode"> </p> <p class="sycode"> output </p> standard output <p class="sycode"> </p> <p> Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.</p> <p> There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1?≤?i?≤?n, the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i?-?1 if i?>?1 or to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i?=?n, the cursor appears at the beginning of the string).</p> <p> When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z' follows 'a'). The same holds when he presses the down arrow key.</p> <p> Initially, the text cursor is at position p.</p> <p> Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?</p> <p class="sycode"> </p> <p class="sycode"> Input </p> <p> The first line contains two space-separated integers n (1?≤?n?≤?105) and p (1?≤?p?≤?n), the length of Nam's string and the initial position of the text cursor.</p> <p> The next line contains n lowercase characters of Nam's string.</p> <p class="sycode"> </p> <p class="sycode"> Output </p> <p> Print the minimum number of presses needed to change string into a palindrome.</p> <p class="sycode"> </p> <p class="sycode"> Sample test(s) </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> input </p> <pre style="代码" class="precsshei">8 3aeabcaez
output
Note
A string is a palindrome if it reads the same forward or reversed.
In the sample test, initial Nam's string is: (cursor position is shown bold).
In optimal solution, Nam may do 6 following steps:
The result, , is now a palindrome.
因为没有删除操作,最后的回文串是什么样已经是确定的了, A-->A' 或 A'--->A 或到A和A'的中间值上下移动的次数是一样的,所以没有必要跨越中点
只要计算出在一边的左右移动次数就可以了....
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn=200100;const int INF=0x3f3f3f3f;char str[maxn];char rstr[maxn];int n,p;int change[maxn][2];void getC(){ for(int i=0;i<n a> A' change[i][0]=max(str[i],rstr[i])-min(str[i],rstr[i]); change[i][0]=min(change[i][0],min(str[i],rstr[i])+26-max(str[i],rstr[i])); /// 1: A --> m >n>>p; p--; cin>>str; int tt=n/2; if(n%2==0) tt--; if(p>tt) { reverse(str,str+n); p=n-1-p; } memcpy(rstr,str,sizeof(str)); reverse(rstr,rstr+n); getC(); int temp=0; ///改变字符的花费 for(int i=0;i=0;i--) { if(change[i][0]!=0) { R=i; break; } } int L=-1; for(int i=0;i R if(p>=L&&pR) { cout <br> <br> <p></p> <p> <br> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> D. Valid Sets </p> <p class="sycode"> </p> <p class="sycode"> time limit per test </p> 1 second <p class="sycode"> </p> <p class="sycode"> memory limit per test </p> 256 megabytes <p class="sycode"> </p> <p class="sycode"> input </p> standard input <p class="sycode"> </p> <p class="sycode"> output </p> standard output <p class="sycode"> </p> <p> As you know, an undirected connected graph with n nodes and n?-?1 edges is called a tree. You are given an integer d and a tree consisting of nnodes. Each node i has a value ai associated with it.</p> <p> We call a set S of tree nodes valid if following conditions are satisfied:</p> <ol> <li> S is non-empty.</li> <li> S is connected. In other words, if nodes u and v are in S, then all nodes lying on the simple path between u and v should also be presented in S.</li> <li> .</li> </ol> <p> Your task is to count the number of valid sets. Since the result can be very large, you must print its remainder modulo 1000000007 (109?+?7).</p> <p class="sycode"> </p> <p class="sycode"> Input </p> <p> The first line contains two space-separated integers d (0?≤?d?≤?2000) and n (1?≤?n?≤?2000).</p> <p> The second line contains n space-separated positive integers a1,?a2,?...,?an(1?≤?ai?≤?2000).</p> <p> Then the next n?-?1 line each contain pair of integers u and v (1?≤?u,?v?≤?n) denoting that there is an edge between u and v. It is guaranteed that these edges form a tree.</p> <p class="sycode"> </p> <p class="sycode"> Output </p> <p> Print the number of valid sets modulo 1000000007.</p> <p class="sycode"> </p> <p class="sycode"> Sample test(s) </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> input </p> <pre style="代码" class="precsshei">1 42 1 3 21 21 33 4
output
input
0 31 2 31 22 3
output
input
4 87 8 7 5 4 6 4 101 61 25 81 33 56 73 4
output
41
Note
In the first sample, there are exactly 8 valid sets: {1},?{2},?{3},?{4},?{1,?2},?{1,?3},?{3,?4} and {1,?3,?4}. Set {1,?2,?3,?4} is not valid, because the third condition isn't satisfied. Set {1,?4} satisfies the third condition, but conflicts with the second condition.
树型DP,从每一个节点走只扩展和根节点 root 权值 在 root
如果扩展到某个节点 w[v]==w[root] 则标记一下,不要重复走
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <vector>using namespace std;typedef long long int LL;const int maxn=2222;const LL mod=1000000007;int n,d,root;LL w[maxn];vector<int> g[maxn];bool vis[maxn][maxn];LL dp[maxn];LL dfs(int u,int fa){ dp[u]=1; for(int i=0,sz=g[u].size();i<sz int v="g[u][i];" if continue vis temp="dfs(v,u);" dp return main cin>>d>>n; for(int i=1; i>w[i]; for(int i=0; i<n-1 i int a cin>>a>>b; g[a].push_back(b); g[b].push_back(a); } LL sum=0; for(int i=1; i <br> <br> <p class="sycode"> E. LIS of Sequence </p> <p class="sycode"> </p> <p class="sycode"> time limit per test </p> 2 seconds <p class="sycode"> </p> <p class="sycode"> memory limit per test </p> 256 megabytes <p class="sycode"> </p> <p class="sycode"> input </p> standard input <p class="sycode"> </p> <p class="sycode"> output </p> standard output <p class="sycode"> </p> <p> The next "Data Structures and Algorithms" lesson will be about Longest Increasing Subsequence (LIS for short) of a sequence. For better understanding, Nam decided to learn it a few days before the lesson.</p> <p> Nam created a sequence a consisting of n (1?≤?n?≤?105) elements a1,?a2,?...,?an (1?≤?ai?≤?105). A subsequence ai1,?ai2,?...,?aik where 1?≤?i1?<?i2 ?<?...?<?ik?≤?n is called increasing if ai1?<?ai2?<?ai3?<?...?<?aik. An increasing subsequence is called longest if it has maximum length among all increasing subsequences.</p> </p> <p> Nam realizes that a sequence may have several longest increasing subsequences. Hence, he divides all indexes i (1?≤?i?≤?n), into three groups:</p> <ol> <li> group of all i such that ai belongs to no longest increasing subsequences.</li> <li> group of all i such that ai belongs to at least one but not every longest increasing subsequence.</li> <li> group of all i such that ai belongs to every longest increasing subsequence.</li> </ol> <p> Since the number of longest increasing subsequences of a may be very large, categorizing process is very difficult. Your task is to help him finish this job.</p> <p class="sycode"> </p> <p class="sycode"> Input </p> <p> The first line contains the single integer n (1?≤?n?≤?105) denoting the number of elements of sequence a.</p> <p> The second line contains n space-separated integers a1,?a2,?...,?an (1?≤?ai?≤?105).</p> <p class="sycode"> </p> <p class="sycode"> Output </p> <p> Print a string consisting of n characters. i-th character should be '1', '2' or '3' depending on which group among listed above index i belongs to.</p> <p class="sycode"> </p> <p class="sycode"> Sample test(s) </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> input </p> <pre style="代码" class="precsshei">14
output
input
41 3 2 5
output
3223
input
41 5 2 3
output
3133
Note
In the second sample, sequence a consists of 4 elements: {a1,?a2,?a3,?a4} = {1,?3,?2,?5}. Sequence a has exactly 2 longest increasing subsequences of length 3, they are {a1,?a2,?a4} = {1,?3,?5} and {a1,?a3,?a4} = {1,?2,?5}.
In the third sample, sequence a consists of 4 elements: {a1,?a2,?a3,?a4} = {1,?5,?2,?3}. Sequence a have exactly 1 longest increasing subsequence of length 3, that is {a1,?a3,?a4} = {1,?2,?3}.
Solution 2:
// Some notation is re-defined.
Let F1i be the length of LIS ending exactly at ai of sequence {a1,?a2,?...,?ai}.
Let F2i be the length of LIS beginning exactly at ai of sequence {ai,?ai?+?1,?...,?an}.
l = length of LIS of {a1,?a2,?...,?an} = max{F1i} = max{F2j}.
Let Fi be the length of LIS of sequence {a1,?a2,?...,?ai?-?1,?ai?+?1,?...,?an} (i.e the length of LIS of initial sequence a after removing element ai).
Index i must in group:
1) if F1i?+?F2i?-?1?
2) if Fi?=?l
3) if Fi?=?l?-?1
How to caculate Fi? We have: Fi?=?max{F1u?+?F2v} among 1?≤?u?
正反求两遍LIS,比较一下即可.....
如果F1[i]+F2[j]-1==LIS 要用map记录下有没有相同的F1[i],F2[i] 有输出2 没有输出3
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <set>#include <map>using namespace std;const int maxn=100100;int n,a[maxn],b[maxn];int f1[maxn],f2[maxn];int v1[maxn],n1,v2[maxn],n2;set<int> st;map<pair>,int> mp;int r[maxn],rn;int ans[maxn];int main(){ scanf("%d",&n); for(int i=0;i<n scanf r sort rn="unique(r,r+rn)-r;" for i="0;i<n;i++)" int id="lower_bound(r,r+rn,a[i])-r;" b lis="1;" if v1 v2 f1 else p1="lower_bound(v1,v1+n1,a[i])-v1;" n1 p2="lower_bound(v2,v2+n2,b[i])-v2;" n2 f2 x="i,y=n-1-i;" ans mp printf putchar return> <br> <br> <p></p> <p><br> </p> <p><br> </p> <p><br> </p> </n></pair></int></map></set></vector></algorithm></cstring></cstdio></iostream>

热AI工具

Undresser.AI Undress
人工智能驱动的应用程序,用于创建逼真的裸体照片

AI Clothes Remover
用于从照片中去除衣服的在线人工智能工具。

Undress AI Tool
免费脱衣服图片

Clothoff.io
AI脱衣机

AI Hentai Generator
免费生成ai无尽的。

热门文章

热工具

记事本++7.3.1
好用且免费的代码编辑器

SublimeText3汉化版
中文版,非常好用

禅工作室 13.0.1
功能强大的PHP集成开发环境

Dreamweaver CS6
视觉化网页开发工具

SublimeText3 Mac版
神级代码编辑软件(SublimeText3)

热门话题

本文讨论了HTML&lt; Progress&gt;元素,其目的,样式和与&lt; meter&gt;元素。主要重点是使用&lt; progress&gt;为了完成任务和LT;仪表&gt;对于stati

本文讨论了html&lt; datalist&gt;元素,通过提供自动完整建议,改善用户体验并减少错误来增强表格。Character计数:159

本文讨论了HTML&lt; meter&gt;元素,用于在一个范围内显示标量或分数值及其在Web开发中的常见应用。它区分了&lt; meter&gt;从&lt; progress&gt;和前

本文讨论了视口元标签,这对于移动设备上的响应式Web设计至关重要。它解释了如何正确使用确保最佳的内容缩放和用户交互,而滥用可能会导致设计和可访问性问题。

HTML适合初学者学习,因为它简单易学且能快速看到成果。1)HTML的学习曲线平缓,易于上手。2)只需掌握基本标签即可开始创建网页。3)灵活性高,可与CSS和JavaScript结合使用。4)丰富的学习资源和现代工具支持学习过程。

本文讨论了&lt; iframe&gt;将外部内容嵌入网页,其常见用途,安全风险以及诸如对象标签和API等替代方案的目的。

HTML定义网页结构,CSS负责样式和布局,JavaScript赋予动态交互。三者在网页开发中各司其职,共同构建丰富多彩的网站。

AnexampleOfAstartingTaginHtmlis,beginSaparagraph.startingTagSareEssentialInhtmlastheyInitiateEllements,defiteTheeTheErtypes,andarecrucialforsstructuringwebpages wepages webpages andConstructingthedom。
