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 Functionstandard input/output1 s, 256 MB	x4256
B	OR in Matrixstandard input/output1 s, 256 MB	x2520
C	Palindrome Transformationstandard input/output1 s, 256 MB	x1769
D	Valid Setsstandard input/output1 s, 256 MB	x643
E	LIS of Sequencestandard input/output2 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

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

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;
}

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:

$第二十一次codeforces竞技结束 #277 Div 2$ where $第二十一次codeforces竞技结束 #277 Div 2$ 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:

$第二十一次codeforces竞技结束 #277 Div 2$ .

(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 2
1 0
0 0

output

NO

input

2 3
1 1 1
1 1 1

output

YES
1 1 1
1 1 1

input

2 3
0 1 0
1 1 1

output

YES
0 0 0
0 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;
}

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第二十一次codeforces竞技结束 #277 Div 2

A. Calculating Function

Code：

B. OR in Matrix

Code：

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