Problem link

I have done this problem with time complexity O( nlogn + n * m ) and space complexity O( 2 * m ). How can i improve this?

Problem link

Why is the answer 2 ^ ( n — 1 ) ?

I have encountered weird behaviour with character arrays/strings on various websites. For example take these two submissions — First submission Second submission. I have just removed fflush(stdin) from my first submission and got AC. I faced similar issue with this problem. This submission has scanf without fflush(stdin) and it got runtime error but according to previous logic it should have got AC. Now this submission with fflush(stdin) and scanf also got runtime error. But this submission with cin got AC.

So my question is what should i do to prevent such nasty errors during contest? cin is slower than scanf so i prefer scanf.

EDIT — For first problem, i saw that i have used different versions of C++. In C++ 11 it gave WA but in C++ it gave AC.

Many people might want to participate in both contests so please look into this issue. 101 Hack Nov and Codeforces Round #332 (Div. 2)

EDIT — Codeforces has shifted the Round #322 to 20th November. Thank you Codeforces. :)

How to take fast I/O in C++? I am solving a problem where number of test cases can be 10^6 so i guess fast I/O will help to speed up my solution.

I want to know how can i compare two different outputs so that i can check whether an algorithm works fine for corner cases also. I want to compare output generated by naive algorithm and an another algorithm saved in two .txt files. I use Code Blocks.

I tried to solve FACTMUL and got AC with 4.82 time. There are many submissions with less than 1.0 time. How to do it in less than 1.0 time?

My code

#include<bits/stdc++.h>
unsigned long long mod=109546051211,fact[587117]={0,1,2,6},a[587117]={0,1,2,12},i,j;
unsigned long long mulmod(unsigned long long a,unsigned long long b)
{
    if((a==0)||(b<mod/a)) return (a*b)%mod;
    unsigned long long x=0,y=a%mod;
    while(b>0)
    {
        if(b%2==1) x=(x+y)%mod;
        y=(y*2)%mod;
        b/=2;
    }
    return x%mod;
}
int main()
{
    for(j=2;j<587117;j++)  //Computes and stores factorial
    {                      //of all number till 587116
        if(fact[j]) continue;
        fact[j]=mulmod(j,fact[j-1]);
    }
    unsigned long long n;
    scanf("%llu",&n);
    if(n>=587117) printf("0\n");
    else
    {
        if(a[n]!=0) printf("%llu\n",a[n]); //It will make no difference
        else                               //because test case is 1
        {
            for(i=2;i<=n;i++) a[i]=mulmod(fact[i],a[i-1]);
            printf("%llu\n",a[n]);
        }
    }
    return 0;
}

Let a = ( 2 ^ 9 ) * ( 3 ^ 2 ) * ( 7 ^ 1 ) * ( 11 ^ 1 )
and b = ( 2 ^ 2 ) * ( 3 ^ 4 ) * ( 7 ^ 1 ) * ( 13 ^ 2 )

then our answer will be ( 2 ^ 9 ) * ( 11 ^ 1 )

I can easily which factors will be present in the answer by calculating a / gcd(a,b) but i am having difficulty in finding the power of those factors. Is it possible to solve the problem using gcd and lcm? Or we will have to factorize both a and b then find the answer?

EDIT — Or is it possible to just find the factors having the exact same power? For above example it is ( 7 ^ 1 ).

For a = ( 2 ^ 3 ) * ( 3 ^ 2 ) * ( 5 ^ 3 ) * ( 7 ^ 1 )
and b = ( 2 ^ 4 ) * ( 3 ^ 2 ) * ( 7 ^ 1 ) * ( 17 ^ 2 )
it is ( 3 ^ 2 ) * ( 7 ^ 1 )

Hi i am trying to solve above problem and getting WA. Help.

#include<stdio.h>
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        char a[1000];
        int i,b=0,c=0,d=0;
        fflush(stdin);
        gets(a);
        fflush(stdin);
        gets(a);
        for(i=0;a[i]!=' ';i++)
        {
            if(a[i]=='m')
            {
                b=-1;
                break;
            }
            b=b*10+(a[i]-'0');
        }
        for(;a[i]!=' ';i++);
        i+=3;
        for(;a[i]!=' ';i++)
        {
            if(a[i]=='m')
            {
                c=-1;
                break;
            }
            c=c*10+(a[i]-'0');
        }
        for(;a[i]!=' ';i++);
        i+=3;
        for(;a[i]!='\0';i++)
        {
            if(a[i]=='m')
            {
                d=-1;
                break;
            }
            d=d*10+(a[i]-'0');
        }
        if(b==-1) printf("%d + %d = %d\n",d-c,c,d);
        else if(c==-1) printf("%d + %d = %d\n",b,d-b,d);
        else printf("%d + %d = %d\n",b,c,b+c);
    }
    return 0;
}

PS — I got AC. I was getting WA because there were many blank lines in input.