6kyu Steps in k-primes

題目背景：

A natural number is called k-prime if it has exactly k prime factors, counted with multiplicity.

Task:

We will write a function kprimes_step(k, step, start, nd) with parameters:

• k (integer > 0) which indicates the type of k-primes we are looking for,
• step (integer > 0) which indicates the step we want to find between two k-primes,
• start (integer >= 0) which gives the start of the search (start inclusive),
• nd (integer >= start) which gives the end of the search (nd inclusive)

題目分析：

本道題主要是 如何計數(shù)素因子的個數(shù)，關(guān)于計數(shù)素數(shù)因子個數(shù)的方法，存在非常經(jīng)典的因子分解的模板思路，先附上計數(shù)素數(shù)個 數(shù)的函數(shù)：

int KPrimes::count_primes(long long num){long long i = 2;int cnt = 0;while( i * i <= num ) {while( num % i == 0 ) {num /= i;cnt++;}i++;}if ( num != 1 ) cnt++;return cnt; }

最終AC的代碼：

#include <vector>namespace KStep{int count_primes(long long num){long long i = 2;int cnt = 0;while( i * i <= num ) {while( num % i == 0 ) {num /= i;cnt++;}i++;}if ( num != 1 ) cnt++;return cnt;}std::vector<std::pair <long, long>> kprimesStep(int k, int step, long long m, long long n){if ( k < 1 || m > n || step < 1 ) return {};std::vector<std::pair <long, long>> res;for ( long long i = m; i <= n - step; i++) {if ( count_primes(i) == k && count_primes(i + step) == k) {std::pair <long, long> seg = std::make_pair ( i, i + step );res.push_back(seg);}}return res;} }

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