10302 - Summation of Polynomials

[mido]

I get WA with this code...any tips :
[cpp]
#include <iostream.h>
#include <stdio.h>
#include <math.h>

int
num,
i;

float sum;

void main()
{
while (!cin.eof() && cin>>num)
{
sum=0;
for (i=0;i<=num;i++)
sum+=pow(i,3);
printf("%.0lf\n",sum);
}
}
[/cpp]

[10153EN]

Clearly, there should be value overflow~

[mido]

Well a friend just mentioned that any number type can only be accurate upto some length after which there will be some zeroes inputted automatically. Oh well.

[Joe Smith]

Well a friend just mentioned that any number type can only be accurate upto some length after which there will be some zeroes inputted automatically. Oh well.

gcc (which is what UVA uses) has a "long long" data type which supports integers up to 2^64. So does Java's "long".

[popel]

1^3 + 2^3 + 3^3 + ... ... ... +n^3

Sn=(n*(n+1)/2)^2

[scythe]

I worked on the helper formulas given in the task and i came up with the following formula:

Sn = ((N-1)*N*(N+1)*(N+2) + 2*N*(N+1)) / 4
which is actually an uglier form of your formula

I got AC using big numbers.

[supermin]

I think this question can just use the long double.=>(%Lf)

it can solve it very quickly.

[Larry]

long long works, and also gets you AC with good time.. =)

[mido]

Anything wrong with this:

[cpp]
#include <iostream.h>
#include <stdio.h>
#include <math.h>

long long
num,
i,j;

long long sum;

void main()
{
while (!cin.eof() && cin>>num && !cin.fail())
{
sum=0;
for (i=0;i<=num;i++)
{
long temp=1;
for (j=0;j<3;j++)
temp*=i;
sum+=temp;
}
cout<<sum<<endl;
}
}


[/cpp]

[shamim]

Your program seems to run ok.

But you might consider using a formula for the sum of Nth cube instead of using the iteration:

for (i=0;i<=num;i++)
{
long temp=1;
for (j=0;j<3;j++)
temp*=i;
sum+=temp;
}

the formula is :

(N*(N+1)/2)^2

[Chow Wai Chung]

This problem seem very easy, and i use (n*(n+1)/2)^2 to calculate the answer, but i get a WA....

Would anyone tell me what is my mistake??

[c]
#include <stdio.h>
#include <math.h>

int main()
{
long long result,n;
while(scanf("%lld",&n)!=EOF)
{
result=(long long)pow(n*(n+1)/2,2);
printf("%lld\n",result);
}
return 0;
}
[/c]

[turuthok]

Your formula looks great and you might want to try x = 50,000 and check your result using calculator. Most likely it is an overflow or precision problem.

-turuthok-

[IIUC GOLD]

Use the following

result = n * n * (n + 1) * (n + 1) /4;

instead of

result=(long long)pow(n*(n+1)/2,2);

[Chow Wai Chung]

Thank you to turuthok and IIUC GOLD

I solved this problem by using n * n * (n + 1) * (n + 1) /4, may be i lost something when cast the pow() to long long before.

[mido]

Got AC long time ago...who would've thought you needed for the temp variable....