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g152xx Group Title

Find the no of positive integral solutions for the equations (1/x) + (1/y) = 1/N! (read 1 by n factorial) Print a single integer which is the no of positive integral solutions modulo 1000007. Input: N Output: Number of positive integral solutions for (x,y) modulo 1000007 Constraints: 1 <= N <= 10^6 Sample Input00: 1 Sample Output00: 1 Sample Input01: 32327 Sample Output 01: 656502 Sample Input02: 40921 Sample Output 02: 686720

  • one year ago
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  1. g152xx Group Title
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    Sample Problem Statement: Given an integer N, print 'hello world' N times. Sample Input 5 Sample Output hello world hello world hello world hello world hello world Solution in C _________________ #include <stdio.h> int main() { int i, n; scanf("%d", &n); for (i=0; i<n; i++) { printf("hello world\n"); } return 0; }

    • one year ago
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    Anyone?

    • one year ago
  3. mayankdevnani Group Title
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    http://sharathpan.wordpress.com/2012/05/17/interview-street-equations-solution/

    • one year ago
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    No of solutions for the equation XY=N! is (e1+1)(e2+1)………..(en+1) where e1,e2…en are multiplicities of Prime Numbers below N. For Ex: XY=4! No of primes below 4= {2,3} 4!= 24= 23*31 where 3 and 1 are prime multiplicities of 24. so applying the formula (3+1)*(1+1)=8 (no of factors of 24={1,2,3,4,6,8,12,24}. Now the equation 1⁄x+1⁄y= N! can be transformed into (x-N!)(y-N!)=N!2 Hence No. of solutions of the above equation is (2e1+1)(2e2+1)………..(2en+1) Hence to solve the problem: 1) First find out the primes less than N 2) Find the prime multiplicities 3) Apply the formula.

    • one year ago
  5. g152xx Group Title
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    thanks @mayankdevnani

    • one year ago
  6. mayankdevnani Group Title
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    welcome

    • one year ago
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