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 #include #include #include #include using namespace std;typedef long long int LL;LL n;int main(){    cin>>n;    if(n%2==0)    {        LL t=n/2;        cout       

                      
   
    
     
      
       
        B. OR in Matrix       
       
        
         time limit per test        
 1 second              
        
         memory limit per test        
 256 megabytes              
        
         input        
 standard input              
        
         output        
 standard output                   
       
 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:
       
  where  is equal to 1 if some ai?=?1, otherwise it is equal to 0.
       
 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:
       
 .
       
 (Bij is OR of all elements in row i and column j of matrix A)
       
 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.
            
       
        Input       
       
 The first line contains two integer m and n (1?≤?m,?n?≤?100), number of rows and number of columns of matrices respectively.
       
 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).
            
       
        Output       
       
 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.
            
       
        Sample test(s)       
       
        
         
          input         
         
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 #include #include #include 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>n>>m;    for(int i=0;i>B[i][j];    for(int i=0;i

        
       
        C. Palindrome Transformation       
       
        
         time limit per test        
 1 second              
        
         memory limit per test        
 256 megabytes              
        
         input        
 standard input              
        
         output        
 standard output                   
       
 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.
       
 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).
       
 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.
       
 Initially, the text cursor is at position 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?
            
       
        Input       
       
 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.
       
 The next line contains n lowercase characters of Nam's string.
            
       
        Output       
       
 Print the minimum number of presses needed to change string into a palindrome.
            
       
        Sample test(s)       
       
        
         
          input         
         
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 #include #include #include #include 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 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       

       
 
                      
   
    
     
      
       
        D. Valid Sets       
       
        
         time limit per test        
 1 second              
        
         memory limit per test        
 256 megabytes              
        
         input        
 standard input              
        
         output        
 standard output                   
       
 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.
       
 We call a set S of tree nodes valid if following conditions are satisfied:
       

 S is non-empty.
        
 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.
        
 .
       
 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).
            
       
        Input       
       
 The first line contains two space-separated integers d (0?≤?d?≤?2000) and n (1?≤?n?≤?2000).
       
 The second line contains n space-separated positive integers a1,?a2,?...,?an(1?≤?ai?≤?2000).
       
 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.
            
       
        Output       
       
 Print the number of valid sets modulo 1000000007.
            
       
        Sample test(s)       
       
        
         
          input         
         
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 #include #include #include #include using namespace std;typedef long long int LL;const int maxn=2222;const LL mod=1000000007;int n,d,root;LL w[maxn];vector 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>d>>n;    for(int i=1; i>w[i];    for(int i=0; i>a>>b;        g[a].push_back(b);        g[b].push_back(a);    }    LL sum=0;    for(int i=1; i        

        E. LIS of Sequence       
       
        
         time limit per test        
 2 seconds              
        
         memory limit per test        
 256 megabytes              
        
         input        
 standard input              
        
         output        
 standard output                   
       
 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.
       
 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?
 Nam realizes that a sequence may have several longest increasing subsequences. Hence, he divides all indexes i (1?≤?i?≤?n), into three groups:
       

 group of all i such that ai belongs to no longest increasing subsequences.
        
 group of all i such that ai belongs to at least one but not every longest increasing subsequence.
        
 group of all i such that ai belongs to every longest increasing subsequence.
       
 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.
            
       
        Input       
       
 The first line contains the single integer n (1?≤?n?≤?105) denoting the number of elements of sequence a.
       
 The second line contains n space-separated integers a1,?a2,?...,?an (1?≤?ai?≤?105).
            
       
        Output       
       
 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.
            
       
        Sample test(s)       
       
        
         
          input         
         
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 #include #include #include #include #include #include