How to calculate sum of the given series - c

Question

The nth term of the series is f(n).
f(n) = 1, when n=1
f(n)= ((n-1)*(8*(n–2)* 2+ 20)+4) , when n>1
P(n)=f(1)+f(2)+.....f(n)
1<=n<=10^9
for a given n we have to find P(n) modulo 10^9+7.
I solved the equation and finally getting the answer as
P(n)=(16n^3-18n^2+14n-12)/3
Problem comes when I implemented it in c++.
Given below the code tell me whats wrong with it and how to resolve it?
#include<stdio.h>
#define c 1000000007
int main()
{
long long int t,n,sum;
scanf("%lld",&t);
while(t--)
{
sum=0;
scanf("%lld",&n);
if(n==1)
printf("1\n");
else
{
sum=(((16*(((((n%c)*(n%c))%c)*(n%c))%c)%c)+14*(n%c))%c-(18*(((n%c)*(n%c))%c)%c+12)%c)/3;
sum++;
printf("%lld\n",sum);
}
}
return 0;
}

Answer 1

There are multiple issues in your code.
Problem 1:
An expression like long long b = ((a%c)*(a%c))%c; with a and b being long long
and c just a "plain number", ie. int, is still subject to int overflow, without
using the full capabilities of long long. The reason is the result of modulo:
long long = ((long long % int) * (long long % int)) % int;
long long = (int * int) % int;
long long = int % int; //overflow here
long long = int;
long long = long long; //expanding too late
//finished
As an example to check (http://ideone.com/kIyM7K):
#include <iostream>
using namespace std;
int main() {
int i = 1000000;
long long ll = 1000000;
cout<< ((999999%i)*(999999%i)*(999999%i)) << endl;
cout<< ((999999%ll)*(999999%ll)*(999999%ll)) << endl;
return 0;
}
In your code, 1000000007 is not a long long ...
Use 1000000007ULL to treat is as unsigned long long
Problem 2:
You´re using (or at least "used", before the edits) %lld in printf and scanf
for unsigned long long, which is %llu. %lld would be a signed long long
Problem 3:
P(n) purely mathematical and without modulo is positive for every natural number n>=1.
But P(0) is negative, and more important: The four parts of your equation modul´t
and the added/subtracted together could result in a negative number, which is congruent
to the positive expected result, but C doesn´t know that.
So, use signed instead of unsigned, and after the whole calculation, check
if the result is <0 and add c to it to make it >0
Problem 4:
A problem with parenthesis and operator precedence somewhere in your long formula.
A working example, but without the loops and input:
#include<stdio.h>
int main()
{
signed long long n, res, c;
n = 2456ULL;
c = 1000000007ULL;
signed long long part1 = (16*(((n%c)*(n%c)%c)*(n%c)%c))%c;
signed long long part2 = (18*((n%c)*(n%c)%c))%c;
signed long long part3 = (14*(n%c))%c;
res = ((part1 - part2 + part3 - 12)%c)/3 + 1;
printf("%lld\n", res);
return 0;
}
Not a problem:
You don´t need a special check for n=1, because the
whole calculation will result in 1 anyways.