爲什麼這個C++ prime checker和summer會返回錯誤的結果？

Question

我不明白爲什麼總和不能正常工作，但總是一致。 它dosent拋出任何錯誤，但從來沒有得到正確的結果。根據wolfram | alpha，我可以在大型計算中關閉多達200萬個。我不知道爲什麼會發生這種情況，所以您可能有任何想法都非常感謝！ 注意：編譯-O3和-std = C++ 14標誌。

代碼：

#include "stdafx.h" 
#include <math.h> 
#include <vector> 
#include <stdio.h> 
#include <iostream> 
typedef unsigned long long ul; 
const ul PRIMES = 1000000; 
bool isPrime(ul n) 
{ 
    if (n <= 1) return false; 
    double sqN = sqrt(n); 
    for (ul i = 3; i <= sqN; i++) { 
     if ((int)n % i == 0) return false; 
    } return true; 
} 
int main() 
{ 
    std::vector<ul> primes; 
    ul sumPrimes = 0; 
    ul numPrimes = 0; 
    for (ul n = 2; n <= PRIMES; n++) if (isPrime(n)) primes.push_back(n); 
    numPrimes = primes.size(); 
    for (ul sp : primes) sumPrimes += sp; 
    std::vector<ul> fizz, buzz, fizzbuzz; 
    ul sumF = 0, sumB = 0, sumFB = 0; 
    ul numF = 0, numB = 0, numFB = 0; 
    for (ul prime = 0; prime < primes.size(); prime++) { 
     if (prime % 15 == 0) { 
      fizzbuzz.push_back(primes[prime]); 
     } 
     else if (prime % 5 == 0) { 
      buzz.push_back(primes[prime]); 
     } 
     else if (prime % 3 == 0) { 
      fizz.push_back(primes[prime]); 
     } 
    } 
    for (ul fb : fizzbuzz) sumFB += fb; 
    for (ul f : fizz) sumF += f; 
    for (ul b : buzz) sumB += b; 
    numF = fizz.size(); numB = buzz.size(); numFB = fizzbuzz.size(); 
    std::cout << "Stats for primes upto\t" << PRIMES << "\n"; 
    std::cout << "Primecount:\t\t" << numPrimes << "\n"; 
    std::cout << "Sum Primes:\t\t" << sumPrimes << "\n"; 
    std::cout << "Fizzcount:\t\t" << numF << "\n"; 
    std::cout << "Sum Fizz:\t\t" << sumF << "\n"; 
    std::cout << "Buzzcount:\t\t" << numB << "\n"; 
    std::cout << "Sum Buzz:\t\t" << sumB << "\n"; 
    std::cout << "FizzBuzzcount:\t\t" << numFB << "\n"; 
    std::cout << "Sum FizzBuzz:\t\t" << sumFB << "\n"; 
    std::system("pause"); 
    return 0; 
} 

這是輸出我得到： output

Answer 1

在你的isPrime方法中，你開始與我= 3.啓動其中i = 2，讓你卸下有2.

bool isPrime(ul n) 
{ 
    if (n <= 1) return false; 
    double sqN = sqrt(n); 
    for (ul i = 3; i <= sqN; i++) { 
     if ((int)n % i == 0) return false; 
    } return true; 
} 

一個因素非質數嘗試

bool isPrime(ul n) 
{ 
    if (n <= 1) return false; 
    double sqN = sqrt(n); 
    for (ul i = 2; i <= sqN; i++) { 
     if ((int)n % i == 0) return false; 
    } return true; 
} 

除非我誤解了一些東西。