第二十一次codeforces竞技结束 #277 Div 2

寝室外面有个男生在和女生打分手电话……时长85分钟，我只能含泪努力，结果还是坑掉了这场……55555‘

Problems

 

 

#Name  A

Calculating Function

standard input/output

1 s, 256 MB

  x4256B

OR in Matrix

standard input/output

1 s, 256 MB

  x2520C

Palindrome Transformation

standard input/output

1 s, 256 MB

  x1769D

Valid Sets

standard input/output

1 s, 256 MB

  x643E

LIS of Sequence

standard input/output

2 s, 256 MB

  x381

大家都表示简单的一场，我一直被打断一直低级错误，WA数空前新高……哭

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

4

output

2

input

5

output

-3

Note

f(4) =  - 1 + 2 - 3 + 4 = 2

f(5) =  - 1 + 2 - 3 + 4 - 5 =  - 3

题意的话就是说，-+-+-+-+这样的加减式子，有n个数字组成，问结果。

分奇偶然后两两一组求和~

Code：

#include <cmath> #include <cctype>#include <cstdio>#include <string>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))bool cmp(const int a, const int b){return a > b;}int main(){long long n=0;cin>>n;if(n%2==0)cout<<n/2;else cout<<n/2-n;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 printm 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

有一个矩阵A，是由01组成的，然后逐行逐列进行OR逻辑运算，得到的结果为B矩阵。

简单来说，就是如果A里有个1，那么B里这个1的所在行和所在列都成1了。

给结果矩阵问原矩阵是什么。

先逐行读，找出全为1的行，记录位置，然后逐列读，找出全为1的列，同时判断不为全1行却出现1的话直接判NO。

获得行列位置之后一一对应，行、列用剩下的都和对方的第一个下标挤一挤就好。

Code：

#include <cmath> #include <cctype>#include <cstdio>#include <string>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;int mat[111][111]={0};int ans[111][111]={0};int visrow[111]={0};int col[111]={0},vcol=0;int row[111]={0},vrow=0;#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))bool cmp(const int a, const int b){return a > b;}int main(){int n,m;cin>>n>>m;int flag=0,zero=0;for(int i=0;i<n;i++){scanf("%d",&mat[i][0]);flag=0;if(mat[i][0]) flag=1,zero=1;for(int j=1;j<m;j++){scanf("%d",&mat[i][j]);if(mat[i][j]) zero=1;if(mat[i][j]!=mat[i][0]){flag=0;}}if(flag) row[vrow++]=i, visrow[i]=1;}for(int j=0;j<m;j++){flag=0;if(mat[0][j]) flag=1;for(int i=1;i<n;i++){if(mat[i][j]!=mat[0][j]){flag=0;}}if(flag) col[vcol++]=j;else{for(int i=0;i<n;i++){if(mat[i][j] && visrow[i]==0){cout<<"NO";return 0;}}}}if(zero==0){cout<<"YES"<<endl;for(int i=0;i<n;i++){printf("%d",mat[i][0]);for(int j=1;j<m;j++){printf(" %d",mat[i][j]);}printf("\n");}return 0;}if(vcol==0 || vrow==0){cout<<"NO";return 0;}int l=max(vcol,vrow),s=min(vcol,vrow),cntp=0;for(cntp=0;cntp<s;cntp++){ans[row[cntp]][col[cntp]]=1;}if(l==vrow)for(;cntp<l;cntp++){ans[row[cntp]][col[0]]=1;}else for(;cntp<l;cntp++){ans[row[0]][col[cntp]]=1;}printf("YES\n");for(int i=0;i<n;i++){printf("%d",ans[i][0]);for(int j=1;j<m;j++){printf(" %d",ans[i][j]);}printf("\n");}return 0;}