can you please explain the logic of mul3 qstn
Can Some Please help me out
i am getting Runtime Error in Multiple of 3 - MULTHREE
My solution is :
https://www.codechef.com/viewsolution/34113626
Why i am getting run time error in carvans prob??
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int a;
int min,l=1; //l will hold the number of cars with maximum speed
scanf("%d",&min);
for(int i = 1;i<n;i++)
{
scanf("%d",&a);
if(a<min)
{
min=a;
l++;
}
}
printf("%d\n",l);
}
return 0;
}
what if sum>10
u have only done for number from [1,10];
what if d0+d1>10,if d0=7 and d1=9
I figured that there was a pattern. But when I submit, it said it was wrong answer.
import sys
import math
def analyze_this(a_list):
if a_list[1]%2==0:
count=int(a_list[1]/2)
else:
if a_list[0]==a_list[2]:
count=math.floor(a_list[1]/2)
else:
count=math.ceil(a_list[1]/2)
return count
T = int(sys.stdin.readline().strip())
if T>0 and T<=10:
x=0
while x<T:
G=int(sys.stdin.readline().strip())
if G>0 and G<=20000:
y=0
while y<G:
arr=list(map(int,sys.stdin.readline().strip().split(ā ')))
if (arr[0]>0 and arr[0]<=2) and (arr[2]>0 and arr[2]<=2) and (arr[1]>0 and arr[1]<=109):
print(analyze_this(arr))
y+=1
x+=1
Can someone please tell me most effective way for finding factorial of number ā¦since in problem FCTRL , I am unable to find factorial of such large nā¦Help Please.
Donāt find factorial of number, instead just count the number of zeroes
checkout this article : Count trailing zeroes in factorial of a number - GeeksforGeeks
hello admin plz help me with the problem coinflip in dsa learming series contest 1
#include<bits/stdc++.h>
using namespace std;
int main()
{
ios_base::sync_with_stdio(0); cin.tie(0);
long long int t;
cin >> t;
while (tā)
{
long long int G;
cin >> G;
while (Gā)
{
long long int i, j, count = 0, n, q;
cin >> i >> n >> q;
long long int c[n];
for (j = 1; j <= n; j++)
{
c[j] = i;
}
//n coins or n rounds
//value of coins (j)
//state of coins c[j]
for (j = 1; j <= n; j++)
{
if (c[j] % 2 == 0)
{
if ((n - j + 1) % 2 == 0)
{
c[j] = 2;
}
else
{
c[j] = 1;
}
}
else
{
if ((n - j + 1) % 2 == 0)
{
c[j] = 1;
}
else
{
c[j] = 2;
}
}
}
for (j = 1; j <= n; j++)
{
//after all operations
if (c[j] == q)
{
count++;
}
else
{
continue;
}
}
cout << count << " " << endl;
}
}
return 0;
}
my solution is giving TLE.
Can someone please explain to me the problem statement ZCO14003? Iām not getting the question properly
Hi,
Can anyone please help me. What I am doing wrong in my solution (CARVANS)
https://www.codechef.com/viewsolution/35777684
In the third case, car I which is moving at the speed of 4 can go at itās max speed, car II cannot go at itās max speed, car III can move at itās max speed of 1 and car IV and V canāt exceed speed 1.
Hence the ans is 2ā¦car I and car III.
Ques : MULTHREE.
Whatās wrong in this code ?
#include<bits/stdc++.h>
using namespace std;
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin >> t;
while(t--){
long long k,sum,d0,d1,cur,remaining;
cin >> k >> d0 >> d1;
sum = d0+d1;
cur = (d0+d1)%10;
remaining=k-2;
while(remaining>0){
sum+=cur;
cur = (cur+cur)%10;
remaining--;
if(cur==0 || cur==2){
break;
}
}
sum += 20*(remaining/4);
int a[4]={2,4,8,6};
for(int i=0; i<remaining%4; i++){
sum+=a[i];
}
if(k==2){
sum = d0+d1;
}
if(sum%3==0){
cout << "YES\n";
}
else{
cout << "NO\n";
}
}
}
In this case, the pattern of speed will be 8 3 x where x<=3 so first two cars are at their max speed(8 ,3).
for the coin flip problem, can anybody give the exact output?
3
2
1 5 2
2 1001 1
3
1 22 2
1 99 1
2 669 1
4
1 10 1
1 11 2
2 12 1
2 13 2
Can anyone help spotting what is wrong here ? Running well and good but wrong answer when submitting : CodeChef: Practical coding for everyone
Consider the test input:
1
10
10 9 8 7 6 5 4 3 2 1
thaks i get it