3871. GCD Extreme Problem code: GCDEX Given the value of N, you will have to find the value of G. The meaning of G is given in the following code G=0; for(k=i;k< N;k++) for(j=i+1;j<=N;j++) { G+=gcd(k,j); } /*Here gcd() is a function that finds the g
题目: Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N. "Oh
Deciphering Password Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2357 Accepted Submission(s): 670 Problem Description Xiaoming has just come up with a new way for encryption, by calculati