913 - Joana and the Odd Numbers

[laicc86]

after i changed the function to

Code: Select all

long long odd_num(long long num)

the Verdict become CE...
i deleted the odd_num and type into main...
it become AC...
it seems like ACM doesn't accept function!?

and thx anyway Jan~

[Jan]

May be you have done something wrong in the prototype declaration.

[apurba]

Code: Select all


deleted after accepted



[sizogee]

Getting TLE...

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#include <stdio.h>

int main(void) {
  long long x = 1;
  int done = 0;

  while (done != 1) {
    scanf("%ld", &x);
    if (x == 0)
      done = 1;
    else {
      x += 1;
      x /= 2;
      printf("%ld\n", (3 * ((2 * (x * x)) - 3)));
      }
  }
  return 0;
}


99999999 and 1000000001 both work. Entering a non-number value makes it go crazy, but this shouldn't be causing the error, right?

[tajbir2000]

Re: 913 - Joana and the Odd Numbers

Post by sizogee on Wed Sep 10, 2008 3:24 am
Getting TLE...

ur program is not breaking at EOF.Read problem stat. wil u get N=0?????

code>>like this >>>
while(scanf("%lld",&x)==1){
....................................
......................................
printf("%lld\n", (3 * ((2 * (x * x)) - 3)));
}

remember %lld>>>not %ld

i think now u will get AC

[sazzadcsedu]

run sample input.
but show wrong output for big number;
plz someone help.
whats wrong???

Code: Select all

 #include<stdio.h>


 int main()
{
  int n,m,p,num;
 long long sum_tot,sum_even,tot_sum,lst3_sum,lst_trm;
   while(scanf("%d",&num)==1)

{

    n=num-2;
   m=n-1;
   p=m/2;

 sum_tot=n*(n+1)/2;

sum_even=p*(p+1);
tot_sum=sum_tot-sum_even;
                 
 tot_sum*=2;
printf("%lld\n",tot_sum);
 lst_trm=(tot_sum+2*num)-1;   
               
lst3_sum=(3*lst_trm)-6;
printf("%lld\n",lst3_sum);
             

}
    return 0;
  }


[newkid]

@sazzadcsedu

change ur int variables to long long. n*(n+1)/2 will overflow if n is big..

[sazzadcsedu]

still wrong ans!!!
need more help.

[newkid]

did you change this
1.
int n,m,p,num;
to
long long n,m,p,num;

2.
while(scanf("%d",&num)==1)
to
while(scanf("%lld",&num)==1)

3. comment the debug line..
printf("%lld\n",tot_sum);
to
//printf("%lld\n",tot_sum);

i am damn sure u didnt do one of this..

[sazzadcsedu]

yea. ur right.
i forgot to comment the debug line.
but main problem was--> rather than,
sum_tot=n*(n+1)/2;
it should be
sum_tot=(n*(n+1))/2;
however, u r realy a great man.
so dont forget to help me later.

[newkid]

@sazzadcsedu
I think you should remove your code.

cheers

[Bidhan]

AC

[rebecca6212]

can anybody tell me what's wrong with the code?
thank you

Code: Select all

#include <cstdlib>
#include <iostream>

using namespace std;

int main(int argc, char *argv[])
{
    int n,x,y;
    int N[63];
    int m;
   while(1){
    printf("n:");
    scanf("%d",&n);
        while(n%2==1){
            x=n/2;
            m=(1+(2*x)-1)*x+1;
            N[0]=m;
            //printf("m=%d\n",N[0]);
           
            for(int i=0;i<n;i++){
                N[i]=m+2*i;
               // printf("%d\t",N[i]);
                }
               
            break;
           
       
        }
        if(n%2==1){
       
        y=N[n-1]+N[n-2]+N[n-3];
        printf("%d",y);
        printf("\n");}
        else break;

        }
   
    system("PAUSE");
    return EXIT_SUCCESS;
}

[helloneo]

can anybody tell me what's wrong with the code?
thank you

The result may not fit in 32 bit integer..
Use unsigned long long..

Remove your code after AC

[@mjad]

i am not understand how i can solve it?
write me algorithm, or give me short hints to solve this problem