Overflow while doing multiplication

Question

I want to calculate value of (N* N *(N+1)/2) mod M where this N can go upto 10^18 and M is at max up to 10^7. I tried to code it but don't know the reason that why it is overflowing. Here is my code :

In main I do something like this :

long long tt=mulmod(N,N+1,MOD)*InverseEuler(2,MOD);
long long mm=mulmod(tt,N,MOD);

And mulmod function find (A*B)%C . It is as follow :

long long mulmod(long long a,long long b,long long c)
{    
    long long x = 0,y = a%c;
    while(b > 0)
    {
        if(b%2 == 1)
        {
            x = (x+y)%c;
        }
        y = (y*2)%c;
        b /= 2;
    }
    return x%c;
}

Also Inverse Euler is something like this :

long long p(long long n,int m,long long int MOD)
{
    if(m == 0) return 1%MOD;

    long long x = p(n,m/2,MOD);
    if(m%2 == 0)
        return (x*x)%MOD;
    else
        return (((x*x)%MOD)*n)%MOD;
}
long long InverseEuler(int n,int MOD)
{
    return p(n,MOD-2,MOD);
}

Please help me in finding error in this code.

Do you actually need `mulmod`? As `M` is up to 10^7, `((N %M) * (N % M)) % M * ((N + 1) % M)` won't overflow. — justanothercoder, Feb 07 '15 at 04:06
It seems that everything is good, but I'd suggest to change all parameters to `long long` — justanothercoder, Feb 07 '15 at 04:15
@justanothercoder Everything is good means my code is fine ? Or your above comment is fine ? — user3840069, Feb 07 '15 at 04:16
`b=b%c` in first function may be good idea. `return x%c;` is redundant. You should extensively test the first function and convince yourself it is working. You wrote `mulmod` but your later function `p` does not use it: why? — Yakk - Adam Nevraumont, Feb 07 '15 at 04:19
@user3840069 https://en.wikipedia.org/wiki/Modular_arithmetic#Remainders (see the "The fundamental property of multiplication in modular arithmetic may thus be written" part). — Alok Singhal, Feb 07 '15 at 04:20
@AlokSinghal What are you trying to say ? I know Modular Arithmetic man — user3840069, Feb 07 '15 at 04:21
@user3840069 I am saying that you can implement `mulmod` as in the first comment by @justanothercoder — Alok Singhal, Feb 07 '15 at 04:23
@AlokSinghal Then how I would be dividing it by 2 ? Wont it affect results ? — user3840069, Feb 07 '15 at 04:24
@AlokSinghal Also my mulmod is being taken from link http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=primalityTesting o I don't think it would be wrong — user3840069, Feb 07 '15 at 04:25
@user3840069 you can divide `N` by 2 if `N` is even, or `N+1` by 2 otherwise. You can do that before you call `mulmod`, no? — Alok Singhal, Feb 07 '15 at 04:28
@user3840069 I see. That link specifically talks about cases when `c` is so big that `(a%c)*(b%c)` might overflow. That's not the case in your example. — Alok Singhal, Feb 07 '15 at 04:31

score 0 · Accepted Answer · answered Feb 07 '15 at 10:46

The thing is, every operation will overflow if the right (or wrong) set of operands are provided.

justanothercode has hinted in comments that modulo 10^7 will not overflow a long long (in fact it won't overflow a long)

You need to make use of the identities that

(a+b)%c == ((a%c) + (b%c))%c
(a*b)%c == ((a%c)*(b%c))%c

The mathematical proofs of these identities are trivial.

With values a and b near the limits of what a long long can represent, adding or multiplying them can overflow. However, the expression on the right of these identities use the modulo operator to avoid additions or multiplications that can overflow.

No need to use a loop either.

Overflow while doing multiplication

1 Answers1