Codeforces Round #277 (Div. 2)_html/css_WEB-ITnose

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

#include <iostream>#include <cstdio>#include <cstring>#include <stack>using namespace std;int main(){#ifdef xxz    freopen("in.txt","r",stdin);#endif // xxz long long n;    while(cin>>n)    {        long long sum = 0;        sum += n/2;        if(n%2) sum -= n;        cout<<sum<<endl;    }    return 0;}

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 mrows 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

把A数组全部赋值为1，然后根据B数组再把A数组一些元素赋值为0，最后算A和B比较，注意如果开始把A全部初始化为0，很难算出来（稍微思考就知道了，如果全部赋值为0，那么1的时候你不知道把A的哪些赋值为1，就容易出错）

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;int n , m;int g[1111][1111];int res[1111][1111];bool check_1(int row , int column){    int ok = 0;     for(int k = 0 ; k < n ; k ++)            ok |= res[row][k];    for(int k = 0 ; k < m ; k ++)            ok |= res[k][column] ;   if(ok) return true;    else return false;}int main(){    #ifdef xxz    freopen("in.txt","r",stdin);    #endif    while(~scanf("%d%d",&m,&n))    {        for(int i = 0 ; i < m ; i ++)        {            for(int j = 0 ; j < n ; j ++)            {                scanf("%d",&g[i][j]);            }        }        for(int i = 0 ; i < m ; i ++)        {            for(int j = 0 ; j < n ; j ++)            {                res[i][j] = 1;            }        }              for(int i = 0 ; i < m ; i ++)        {            for(int j = 0 ; j < n ; j ++)            {                if(g[i][j] == 0)                {                    for(int k = 0 ; k < n ; k ++)                        res[i][k] = 0;                    for(int k = 0 ; k < m ; k ++)                        res[k][j] = 0;                }            }        }        int flag1 = 0;        for(int i = 0 ; i < m ; i ++)        {            for(int j = 0 ; j < n ; j ++)            {                if(g[i][j] == 1)                {                    if(!check_1(i , j))                        flag1 = 1;                }                if(flag1)                    break;            }            if(flag1)                break;        }        if(flag1)            printf("NO\n");        else        {            printf("YES\n");            for(int i = 0  ; i < m  ; i ++)            {                for(int j = 0 ; j < n ; j ++)                {                    printf("%d%c",res[i][j], j == n - 1 ? '\n' : ' ');                }            }        }    }}

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.

只要考虑一半就行，然后判断一半的一半离pos是近还是远计算就行

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;int n , pos;char s[100100];int main(){#ifdef xxz    freopen("in.txt","r",stdin);#endif    int T,Case = 0;;    while(cin>>n>>pos)    {      pos--;      cin>>s;      int len = strlen(s);      int l = 0,r = n/2 - 1;      if(n/2 <= pos) pos = n - pos-1;      while(l < n/2 && s[l] == s[n-1-l]) l++;      while(r >=0 && s[r] == s[n-1-r] ) r--;      int ans = 0;      for(int i = l; i <= r; i++)      {        int temp = abs(s[i] - s[n-1-i]);        ans += min(temp,26 - temp);      }      if(l < r)      {        ans += min(abs(pos - l), abs(r - pos)) + r- l;      }      else if(l == r)      {          ans += abs(pos - r);      }      cout<<ans<<endl;    }    return 0;}

D:

树形DP，目前还不会，等打完广州区域赛在学习下树形DP再补上