ZSUM

Just Add It

Link to the question : ZSUM 

HINTS :

When trying to find  (Z_n+Z_n-1-2Z_n-2) you will notice that most of the terms of S_n and P cancel out and the terms left can be easily calculated by modular exponention.   

  You can find the code for modular exponention in my previous post YAPP .

SOURCE CODE :

/* Just Add */
/* Sushant Gupta */

#include<stdio.h>
#define mod 10000007

long long int mult(long long int x,long long int y)
{
    long long int ans =1;
    while(y>0)
    {
        if(y%2)
            ans = (ans * x) % mod;
        x = (x * x) % mod;
        y=y>>1;

    }
    return ans;
}

int main()
{
    long long int n,k;
    while(1)
    {
        scanf("%lld%lld",&n,&k);
        if(n==0 && k==0 )
            return 0;
        else
        {
            long long int s1, s2,s3,s4;
            s1 = (2 * mult(n-1,n-1) ) % mod;
            s2 = (2 * mult(n-1,k) ) % mod;
            s3 = mult(n,k);
            s4 = mult(n,n);
            long long int res = (s1 + s2 +s3 + s4) % mod;
            printf("%lld\n",res);
        }
    }

}

YAPP

Yet Another Permutations Problem

Link to the question : YAPP 

HINT : 

All you need to find is 2 ^ (n-1). Use modular exponention to calculate this. You can read articles  on modular exponention from wikipedia or if possible can understand from my blog.

 SOURCE  CODE :

/* Yet Another Permutation Problem */
/* Sushant Gupta */

#include<stdio.h>


int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long int n,ans=1,mod =  1000000007,a=2;
        scanf("%lld",&n);
        n=n-1;
        if(n==0)
            printf("1\n");
        else
        {
            while(n>0)
            {
                if(n&1)
                    ans = ans * a % mod;
                a = a*a % mod;
                n= n>>1;

            }
            printf("%lld\n",ans);
        }
    }
    return 0;
}

 

MAIN74

Euclids algorithm revisited

Link to the question : MAIN74 

HINT :

The numbers form the fibonacci sequence if you try solving it for some test cases.

Example for n= 2 loops the answer will be 5 (3,2).

              for n= 3 loops, result is 8 (5,3).

              for n=4 loops, output will be 13 (8,5). 

And so on...

Hence it is obivious that for n >=2 output is (n + 3)th fibonacci term.

Now to solve the problem in O(logn) time you should use the optimized matrix multiplication method to generate the fibonacci term.

SOURCE CODE :

#include<stdio.h>

long long MAX=1000000007;

void multiply(long long F[2][2],long long M[2][2]);
void power(long long F[2][2],long long n);
long long fib(long long n);

long long fib(long long n)
{
    long long F[2][2]={{1,1},{1,0}};
    if(n==0)
        return 0;
    power(F,n-1);
    return F[0][0];
}

void multiply(long long F[2][2],long long M[2][2])
{

    long long x = ((F[0][0]*M[0][0])%MAX + (F[0][1]*M[1][0])%MAX)%MAX;
    long long y =  ((F[0][0]*M[0][1])%MAX + (F[0][1]*M[1][1])%MAX)%MAX;
    long long z =  ((F[1][0]*M[0][0])%MAX + (F[1][1]*M[1][0])%MAX)%MAX;
    long long w =  ((F[1][0]*M[0][1])%MAX + (F[1][1]*M[1][1])%MAX)%MAX;

    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;

}

void power(long long F[2][2], long long n)
{
    if( n == 0 || n == 1)
        return;
    long long M[2][2] = {{1,1},{1,0}};

    power(F, n/2);
    multiply(F, F);

    if( n&1)
    multiply(F, M);
}

int main()
{
    int t;
    long long int n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld", &n);
        if(n==0)printf("0\n");
        else if(n==1)printf("2\n");
        else printf("%lld\n",(fib(n+3))%MAX);
    }

    return 0;
}