I am getting this error. I have tried this code for many sample input. However, it is showing NZEC error when i try submitting the code.
MY CODE :
I am getting this error. I have tried this code for many sample input. However, it is showing NZEC error when i try submitting the code.
MY CODE :
I am also having the same problem of nzec in following program.
import java.util.*;
class Solution {
static long want = 0, red = 0;
static int i = 0;
static long[] a= new long[(int)10000];
static long[] b= new long[(int)10000];
static void marks(long e,long n) {
Scanner S = new Scanner(System.in);
if( n == 1) {
while( e > 1 ) {
long sum = S.nextLong();
a[i] = a[i] + sum;
e--;
}
}
else {
while( e > 0 ) {
long sum = S.nextLong();
a[i] = a[i] + sum;
e--;
}
}
i++;
}
static void sort() {
long temp = 0;
for ( int j = 0 ; j < i ; j++ ) {
for( int k = j+1 ; k < i ; k++ ) {
if( a[j] < a[k] ) {
temp = a[k];
a[k] =a[j];
a[j] = temp;
}
}
}
}
static void select(int k) {
int z = 0;
while ( k > 0 ) {
b[z] = a[k-1];
z++;
k--;
}
}
static long min(long k) {
if ( red > b[0] )
want = 0;
else if ( red == b[0] )
want = 1;
else
want = b[0] - red + 1;
return want;
}
public static void main(String[] args) {
Scanner S = new Scanner(System.in);
long t = S.nextLong();
while(t>0) {
for ( int l = 0 ; l < a.length ; l++ ){
a[l] = 0;
b[l] = 0;
}
i = 0;
want = 0;
red = 0;
long n = S.nextLong();
if( n < 1 || n > 10000 )
System.exit(0);
int k = S.nextInt();
if( n < 1 || n > 10000 )
System.exit(0);
long e = S.nextLong();
if( e < 1 || e > 4 )
System.exit(0);
long m = S.nextLong();
if( m < 1 || m > 1000000000 )
System.exit(0);
while( n > 0) {
marks(e,n);
n--;
}
red = a[i-1];
sort();
select(k);
long ans = min(k);
System.out.println(""+ans);
t--;
}
}
}
I am facing NZEC runtime error in this code… what wrong am i doing??
t=int(raw_input())
while(t>0):
c=int(input())
d=int(input())
l=int(input())
if l%4==0:
l1=l/4
if l1<= c+d:
if c<=2d and l1>=d:
print “yes\n”
elif c>2d and l1>= c+d-2*d:
print “yes\n”
else:
print “no\n”
else:
print “no\n”
else:
print “no\n”
t-=1
I highly suspect that the error is due to missing return statement and “main()”
Change it to “int main()” and add a “return 0;” in end of program.
Plz Help me I Am Getting Same Error
Code Here-
/* package codechef; // don’t place package name! */
import java.util.;
import java.lang.;
import java.io.*;
/* Name of the class has to be “Main” only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
BufferedReader p = new BufferedReader(new InputStreamReader(System.in));
int queries= Integer.parseInt(p.readLine());
int big,small;
for (int i =0;i<=queries;i++)
{
int a= Integer.parseInt(p.readLine());
int b= Integer.parseInt(p.readLine());
if(a>b)
{
big=a;
small=b;
}
else
{
big=b;
small=a;
}
if(a%2!=0 && b%2!=0)
{
if(a+2==b || a-2==b)
System.out.println("Yes");
}
else if (a%2==0 && b%2==0)
{
if(a+2==b || a-2==b)
System.out.println("Yes");
}
else if (big-small==1 && big%2==0)
System.out.println("Yes");
else
System.out.println("No");
}
}
}
def home():
x=[]
for number in range(0,4):
value = int(input('Enter the values for %dth element'%number))
x.append(value)
if x[2] - x[0] > 0 and x[1] - x[3] == 0:
print (‘right’)
elif x[0] - x[2] > 0 and x[1] - x[3] == 0:
print(‘left’)
elif x[0]-x[2] == 0 and x[1] - x[3] > 0 :
print ('down')
elif x[0] - x[2] == 0 and x[1] - x[3] > 0 :
print ('up')
else :
print (‘sad’)
test_case= int(input(‘Enter the number of test cases’))
for no in range(0,test_case):
home()
I’m getting NZEC for this code. Someone help me out!
How remove NZEC
/* package codechef; // don’t place package name! */
import java.util.;
import java.lang.;
import java.io.*;
/* Name of the class has to be “Main” only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
int x=sc.nextInt();
int y=sc.nextInt();
if (y>x)
{
double b=(y-x)-0.50;
System.out.println(b);
}
else System.out.println(y);
sc.close();
}
}
why does the following code give NZEC error ?
T = input()
N = [0 for i in range (0,T)]
P = [0 for i in range(T)]
res = [“no” for i in range(T)]
p = [0 for i in range(500)]
cc = [0 for i in range(500)]
ch = [0 for i in range(500)]
for i in range(0,T):
N[i] = input()
P[i] = input()
for j in range (0, N[i]):
p[j] = input()
if (p[j]>=(P[i]/2)):
cc[i] = cc[i]+1
elif (p[j]<=(P[i]/10)):
ch[i] = ch[i]+1
if((cc[i] == 1) and (ch[i] == 2)):
res[i] = “yes”
for i in range(0,T):
print res[i]
[Solution][1]
→ Why does this code gives NZEC??
Thank You
[1]: CodeChef: Practical coding for everyone
adding an exception handler which does nothing removes the nzec error
java example:
try
{
} catch (Exception e){}
actually i use to code in python where there is no such use of return 0 statement.
I have submitted many problems in Python.
Then its due to some exception trace your source code carefully
my code is
import java.io.;
import java.lang.;
public class sum
{
public static void main (String[] args)throws Exception
{
java.io.BufferedReader r = new java.io.BufferedReader (new java.io.InputStreamReader (System.in));
System.out.println(“Enter the number”);
int i=Integer.parseInt(r.readLine());
int sum=0;
for(int j=1;j<i;j++)
{
if(i%j==0)
{
sum+=j;
} }
System.out.println(sum);
}
}
But it is not showing any runtime error! What are you talking about? See-
If n>100 in your code , it will give you NZEC error
you can declare your array z dynamically or declare it with max value of n
Due to brute force consuming lot of memory in recursion calls and overflowing the memory stack. Use map data structure and memoization.
Thank you so much, i tried the memoization and got an AC.
can anyone tell me t=why NZEC error is there
solution
https://www.codechef.com/viewsolution/24654440
q
https://www.codechef.com/JUNE19/PROXYC
When i encountered NZEC problem, I found that I am using arraylist and adding values in it on basis of some condition and when condition is not fulfilled no value will be added in arraylist and in my next statements i was trying to access elements from empty arraylist which result into NZEC error, So while tracing your code keep in mind these things they may help you.