Algorithm

Problem Name: 2 AD-HOC - beecrowd | 1267

Pascal Library

By Ricardo Anido  Brazil

Timelimit: 1

Pascal University, one of the oldest in the country, needs to renovate its Library Building, because after all these centuries the building started to show the eﬀects of supporting the weight of the enormous amount of books it houses.

To help in the renovation, the Alumni Association of the University decided to organize a series of fund-raising dinners, for which all alumni were invited. These events proved to be a huge success and several were organized during the past year. (One of the reasons for the success of this initiative seems to be the fact that students that went through the Pascal system of education have fond memories of that time and would love to see a renovated Pascal Library.)

The organizers maintained a spreadsheet indicating which alumni participated in each dinner. Now they want your help to determine whether any alumnus or alumna took part in all of the dinners.

Input

The input contains several test cases. The ﬁrst line of a test case contains two integers N and D indicating respectively the number of alumni and the number of dinners organized (1 ≤ N ≤ 100 and 1 ≤ D ≤ 500). Alumni are identiﬁed by integers from 1 to N. Each of the next D lines describes the attendees of a dinner, and contains N integers Xi indicating if the alumnus/alumna i attended that dinner (Xi = 1) or not (Xi = 0). The end of input is indicated by N = D = 0.

Output

For each test case in the input your program must produce one line of output, containing either the word ‘yes’, in case there exists at least one alumnus/alumna that attended all dinners, or the word ‘no’ otherwise. The output must be written to standard output.

Alumna: a former female student of a particular school, college or university.
Alumnus: a former male student of a particular school, college or university.
Alumni: former students of either sex of a particular school, college or university.

 Sample Input Sample Output 3 3 1 1 1 0 1 1 1 1 1 7 2 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 yes no

Code Examples

#1 Code Example with C Programming

```Code - C Programming```

``````
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define true 1
#define false 0

int main(int argc, char **argv)
{

int linhas, colunas, i, j;

while (scanf("%d %d", &colunas, &linhas), linhas)
{

char matrix[linhas][colunas];

for (i = 0; i  <  linhas; ++i)
for (j = 0; j  <  colunas; ++j)
scanf("%hhd", &matrix[i][j]);

int cont;
_Bool is = false;
for (j = 0; j  <  colunas; ++j)
{

cont = 0;
for (i = 0; i  <  linhas; ++i)
if (matrix[i][j])
cont++;

if (cont == linhas)
is = true;

}

printf("%s\n", is ? "yes" : "no");

}

return 0;

}
``````
Copy The Code &

Input

cmd
3 3
1 1 1
0 1 1
1 1 1
7 2
1 0 1 0 1 0 1
0 1 0 1 0 1 0
0 0

Output

cmd
yes
no

#2 Code Example with C++ Programming

```Code - C++ Programming```

``````
#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i  < = b; ++i)
#define RFOR(i, b, a) for (int i = b; i >= a; --i)
#define REP(i, N) for (int i = 0; i  <  N; ++i)
#define MAX 1000010
#define pb push_back
#define mp make_pair

using namespace std;

typedef vector<int> vi;
typedef long long int64;
typedef unsigned long long uint64;

int bib[510][110], alumn[110];

int main()
{
int a, b;
while (scanf("%d %d", &a, &b) && a + b) {
REP(i, a)
alumn[i] = 1;
REP(i, b)
{
REP(k, a)
{
scanf("%d", &bib[i][k]);
alumn[k] &= bib[i][k];
}
}
int f = false;
REP(i, a)
{
if (alumn[i] & 1) {
printf("yes\n");
f = !f;
break;
}
}
if (!f)
printf("no\n");
}
return 0;
}
``````
Copy The Code &

Input

cmd
3 3
1 1 1
0 1 1
1 1 1
7 2
1 0 1 0 1 0 1
0 1 0 1 0 1 0
0 0

Output

cmd
yes
no

#3 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const { readFileSync } = require("fs")
const input = readFileSync("/dev/stdin", "utf8").split("\n").map(line => line.split(" "))

const output = []

while (input.length > 0) {
const [A, D] = input.shift()

if (+A == 0 || +D == 0) break
if (isNaN(+A) || isNaN(+D)) break

const aluminisList = new Array(+A).fill(true)
const dinnersList = input.splice(0, +D).map(dinner => dinner.slice(0, +A))

for (const dinner of dinnersList) {
for (let index = 0; index  <  dinner.length; index++) {
if (aluminisList[index] == false) continue
else if (dinner[index] == "0") aluminisList[index] = false
}
}

output.push(aluminisList.includes(true) ? "yes" : "no")
}

console.log(output.join("\n"))
``````
Copy The Code &

Input

cmd
3 3
1 1 1
0 1 1
1 1 1
7 2
1 0 1 0 1 0 1
0 1 0 1 0 1 0
0 0

Output

cmd
yes
no

#4 Code Example with Python Programming

```Code - Python Programming```

``````
while True:
e = str(input()).split()
s = int(e[0])
t = int(e[1])
if s == t == 0: break

m = []
for i in range(t):
m.append([int(x) for x in str(input()).split()])

o = [[lin[j] for lin in m] for j in range(len(m[0]))]

ok = False
for i in range(s):
if o[i].count(0) == 0: ok = True
print('yes' if ok else 'no')
``````
Copy The Code &

Input

cmd
3 3
1 1 1
0 1 1
1 1 1
7 2
1 0 1 0 1 0 1
0 1 0 1 0 1 0
0 0

Output

cmd
yes
no