find nth prime number c

[j]*t); }}} sort (K.begin (), K.end ()); S.push_back (k[0]); mp[k[0]]++; } sort (S.begin (), S.end ()); printf ("The "th ugly number is 859963392."); return 0;}The final output is a well-calculated 1500th result. The input is the size of the ugly number.There is a wrong idea at first to do this question.My answer is to use 2,3,5,6,20,15,30 as the base to multiply the number of not

factorization: at for(intI=2; i*i) -{if(temp%i==0) {s.push_back (i); while(temp%i==0) temp/=i;} } -S.push_back (b);//because B itself may be a prime number, adding B to the vector, even if it is not a prime number, has no effect on the result. - for(intI=0; I//generates an approximate s.size (). -{Su=s[i];

Output768016016 AuthorSmallBeer (CRF) Source hangdian ACM training team training competition (VII) Recommendlcy Question Analysis: If this question is purely violent (it is to use two for loops to find the data scale from 1 ~ The number of elements in I that interact with I), the timeout is very high (n itself is large, and the brute force algorithm> O (n ^ 2) is also u

Title Source: POJ 2992 divisors Test instructions: ... Idea: The uniqueness of prime decomposition a number can be decomposed into several prime numbers multiplied P1^X1*P2^X2*...*PN^XN Number of factors according to multiplication principle (x1+1) * (x2+1) *...* (xn+1) Can not directly

Find the maximum Problem descriptioneuler's totient function, Phi (n) [sometimes called the Phi function], is used to determine the number of numbers less than n which are relatively prime to n. for example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, Phi (9) = 6. Hg is the master of x

Tags: refresh Principle Ultraviolet A 4683 This question refers to a set with a maximum of 12 elements. Find the nth number that can only be divided by one of the sets and only one number. (N I did this by using the refresh principle. First, we will accumulate the number