Can someone find a failing testcase for my sol CodeChef: Practical coding for everyone
It passes the very last test, no idea what is wrong.
Thank you!
good question but poorly explanied by the video guy you should have explained the code along with the cases so it becomes more easy to understand … it looked like you were just trying to wrap it up asap when you were showing the code
You are failing on test case for example
N=5
1 2 3 4 5
Q=5
r=7
the answer should be 2 but your code is showing 1.
So you are failing on all the input where the first element is 1
2 Likes
My whole code failed because of one edge test case alone.I was so frustrated.
#include <bits/stdc++.h>
#define ll unsigned long long int
#define MOD 1000000007
using namespace std;
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin>>t;
while(t–){
ll n;
cin>>n;
ll arr[n];
for(int i=0;i<n;i++) cin>>arr[i];
ll chef[n],chefGreed[n];
memset(chef,0,sizeof(chef));
memset(chefGreed,0,sizeof(chef));
if(arr[0]==1){
for(int i=0;i<n;i++){
chef[i]=1;
if(i==0 && n>1) chefGreed[i]=0;
else chefGreed[i]=1;
}
}
else{
// normal case + greedy case
if(arr[0]%2==0){
if(n>1) chef[0]=arr[0];
else chef[0]=arr[0]-1;
}
else{
if(n>1) chef[0]=arr[0]-1;
else chef[0]=arr[0];
}
chefGreed[0]=arr[0];
for(int i=1;i<n;i++){
if(i==n-1){
if(arr[i]%2==0) chef[i]=(chef[i-1]+arr[i]-1)%MOD;
else chef[i]=(chef[i-1]+arr[i])%MOD;
}
else if(i+1<n-1 && arr[i+1]==1){
if(arr[i]%2==0){
chef[i]=(chef[i-1]+arr[i]-1)%MOD;
chef[i+1]=(chef[i])%MOD;
chefGreed[i]=(chef[i]+1)%MOD;
chefGreed[i+1]=(chefGreed[i]+1)%MOD;
i++;
continue;
}
else{
chef[i]=(chef[i-1]+arr[i])%MOD;
chef[i+1]=(chef[i])%MOD;
chefGreed[i]=(chef[i])%MOD;
chefGreed[i+1]=(chefGreed[i])%MOD;
i++;
continue;
}
}
else{
if(arr[i]%2==0) chef[i]=(chef[i-1]+arr[i])%MOD;
else chef[i]=(chef[i-1]+arr[i]-1)%MOD;
}
chefGreed[i]=(chef[i-1]+arr[i])%MOD;
}
}
// for(int i=0;i<n;i++){
// cout<<chef[i]<<" ";
// }
// cout<<"\n";
// for(int i=0;i<n;i++){
// cout<<chefGreed[i]<<" ";
// }
// cout<<"\n";
int q;
cin>>q;
while(q--){
ll r;
cin>>r;
ll ans=0;
if(r<=n){
if(arr[0]!=1) ans=chefGreed[r-1];
else ans=1;
}
else{
if(r%n==0 || (n==1 && arr[0]==1)){
ans=(ans+((chef[n-1]*((r/n-1)%MOD))%MOD))%MOD;
ans=(ans+chefGreed[n-1])%MOD;
}
else{
ans=(ans+((chef[n-1]*((r/n)%MOD))%MOD))%MOD;
ans=(ans+chefGreed[(r%n)-1])%MOD;
}
// if(r%n==0) ans=(ans+chefGreed[n-1])%MOD;
// else ans=(ans + chefGreed[r%n])%MOD;
}
cout<<(ans%MOD)<<"\n";
}
}
return 0;
}
My code has also passed subtask 2 but it fails for 15 pts. I have tried all test case and inserted MOD everywhere. Can you please tell me where I am wrong?
i have one small doubt, lets say the candy[ ]={2,5,6,3 };
and I am standing at the last index i.e 3,if I choose 3 candies here will my place change 2 times?
For Test case
1
3
3 1 5
1
3
Expected output: 8
Your output: 7
1 Like
Your place will change once but the status of counter will be the same even after swapping places
Both mine and tester’s solution is commented. If you get stuck somewhere, feel free to ask 
https://www.codechef.com/viewsolution/39797040
Please tell any test case where my code fails. I have verified with editorial also and it passes all test cases mentioned there. It would be very helpful if you can suggest some case where my code can get fails
Thanks in Advance
One of the test case where this code fails is
1
2
1 5
1
4
Expected output: 2
anyone plz reply me
I always thought the value could be the same, my goodness.
But I think output should be 6 corresponding to it. Chef will have to take 1 in first turn and then has to swap position with chefu. After third turn chefu and chef again exchange position and chef will take 5 candies finally. So total 6 candies.
Turn 1: Chef takes 1, he has no other option, Switching
Turn 2: Chefu will take 4 here.
end of array so we have to switch again. Chef on open counter.
Turn 3: Chef takes 1 again, Again switch
Turn 4: Chefu takes 4
Ans: 2
I have not looked at your code, but my attempts at a solution also passed only the last test, so I guess we made the same mistake. When the first number in the sequence of A is 1, what does a player do on the last number of each round of N plays? Normally the best thing to do is to take an even number to stay there, so that the other player is forced to take the 1, and so not be able to take any more in that round. But when the total number of rounds R is 1 more than a multiple of N (R % N == 1), the player should take an odd number at the end of the last full round, and then take the final 1.
1 Like
Hi all and @utkarsh911 , Can someone please let me know what is wrong with my submission , I am getting subtask 1 task 0 and subtask 2 and task 2 as wrong answer . I have tried with multiple different testcases and I have also checked for all the testcases provided in the editorial. My solution is here:
https://www.codechef.com/viewsolution/39938632
It will help me a lot to check what went wrong , thanks
I thought I am the biggest fool ,but this comment now is giving a relief :