程序编译但不执行

Question

我编译了程序，但是当我输入=600851475143时没有结果。程序是找到最大的素数（13195的素数是5、7、13和29。最大的素数是多少？号码 600851475143？） 怎么了？

#include<stdio.h>


int isprime(unsigned long  n){
    unsigned long i;
    for(i=2;i<n;i++){
        if((n%i)==0){
            return 0;
        }
    }
    return 1;
}

int main(){
    unsigned long n,i,lpf;
    scanf("%ld",&n);
    for(i=2;i<n;i++){
        if(n%i==0){
            if(isprime(i)==1){
            lpf=i;}
        }
    }
    printf("%ld",lpf);
    return 0;
}

Answer 1

你有一个 for 循环，它运行的次数大约是你输入的数字，i 从 2 到 n-1。在那个 for 循环中，您调用一个函数，该函数大致运行 i 次，这与循环中的 i 相同。这意味着您的程序围绕n^2 进行数学计算，其中n 是您的输入。如果输入一个 12 位数字，n^2 是 24 位数字的大小，对于一个程序来说，做一个 10^24 的计算会花费很多时间。所以你的程序确实运行，它只是计算。

Answer 2

早些时候，我注意到我原来的解决方案：

600851475143 仍然需要很长时间，所以是时候寻找一个 完全不同的算法

搜索 SO 出现 efficient ways of finding the largest prime factor of a number。

这是一个基于@under5hell 在该页面上的回答的简单快速的解决方案：

#include <stdio.h>
#include <assert.h>

int main()
{
    unsigned long long number;

    assert(sizeof(number) * 8 >= 64); // make sure we've enough bits

    scanf("%llu", &number);

    for (unsigned long long divisor = 2; divisor < number; divisor++) {
        if (number % divisor == 0) {
            number /= divisor--; // decrement to remove multiples
        }
    }

    printf("%llu\n", number);

    return 0;
}

哪一个能在几分之一秒内找到 600851475143 的最大素数：

输出

> ./a.out
600851475143
6857
> 

故事的寓意似乎是在查询 SO 之前先搜索 SO！

Answer 3

正如其他人已经说过的：你不需要走完整的路。上到n的平方根就足够了，你也可以在每次找到一个因数的时候减少n本身的值。

一种稍微快一点的方法是使用轮子，尽管时间复杂度与原始版本相同。这里使用的轮子是 11 个粗略数字之间的距离（加上之前的素数和下一轮的偏移量）。

大家一起：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define ISPRIME(x) isprime(x)
//#define ISPRIME(x) isprime_wheel(x)

static int wheel[] = {
1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2,
4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 2, 4, 6, 2, 10, 2, 4, 2, 12, 10, 2, 4, 2,
4, 6, 2, 6, 4, 6, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 6, 8, 6, 10, 2, 4, 6,
2, 6, 6, 4, 2, 4, 6, 2, 6, 4, 2, 6, 10, 2, 10, 2, 4, 2, 4, 6, 8, 4, 2, 4, 12,
2, 6, 4, 2, 6, 4, 6, 12, 2, 4, 2, 4, 8, 6, 4, 6, 2, 4, 6, 2, 6, 10, 2, 4, 6, 2,
6, 4, 2, 4, 2, 10, 2, 10, 2, 4, 6, 6, 2, 6, 6, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8,
4, 2, 6, 4, 8, 6, 4, 6, 2, 4, 6, 8, 6, 4, 2, 10, 2, 6, 4, 2, 4, 2, 10, 2, 10,
2, 4, 2, 4, 8, 6, 4, 2, 4, 6, 6, 2, 6, 4, 8, 4, 6, 8, 4, 2, 4, 2, 4, 8, 6, 4,
6, 6, 6, 2, 6, 6, 4, 2, 4, 6, 2, 6, 4, 2, 4, 2, 10, 2, 10, 2, 6, 4, 6, 2, 6, 4,
2, 4, 6, 6, 8, 4, 2, 6, 10, 8, 4, 2, 4, 2, 4, 8, 10, 6, 2, 4, 8, 6, 6, 4, 2, 4,
6, 2, 6, 4, 6, 2, 10, 2, 10, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 6, 6, 4,
6, 8, 4, 2, 4, 2, 4, 8, 6, 4, 8, 4, 6, 2, 6, 6, 4, 2, 4, 6, 8, 4, 2, 4, 2, 10,
2, 10, 2, 4, 2, 4, 6, 2, 10, 2, 4, 6, 8, 6, 4, 2, 6, 4, 6, 8, 4, 6, 2, 4, 8, 6,
4, 6, 2, 4, 6, 2, 6, 6, 4, 6, 6, 2, 6, 6, 4, 2, 10, 2, 10, 2, 4, 2, 4, 6, 2, 6,
4, 2, 10, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 12, 6, 4, 6, 2, 4, 6, 2, 12,
4, 2, 4, 8, 6, 4, 2, 4, 2, 10, 2, 10, 6, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4,
2, 10, 6, 8, 6, 4, 2, 4, 8, 6, 4, 6, 2, 4, 6, 2, 6, 6, 6, 4, 6, 2, 6, 4, 2, 4,
2, 10, 12, 2, 4, 2, 10, 2, 6, 4, 2, 4, 6, 6, 2, 10, 2, 6, 4, 14, 4, 2, 4, 2, 4,
8, 6, 4, 6, 2, 4, 6, 2, 6, 6, 4, 2, 4, 6, 2, 6, 4, 2, 4, 12, 2, 12
};

int isprime_wheel(unsigned long n)
{
  unsigned long isqrt = (unsigned long) (floor(sqrt(n)) + 1);
  unsigned long start = 5, factor = 2;
  int wlen = sizeof(wheel) / sizeof(int);
  int next = 0;
  if(n == 2){
    return 1;
  }
  if ((n & 1) == 0) {
    return 0;
  }
  if (isqrt * isqrt == n) {
    return 0;
  }
  while (factor < isqrt) {
    if (n % factor == 0) {
      return 0;
    }
    factor += (unsigned long) wheel[next];
    next++;
    if (next == wlen) {
      next = start;
    }
  }

  return 1;
}

unsigned long primefactors(unsigned long n)
{
  unsigned long start = 5, factor = 2;
  unsigned long biggest = 0;
  int wlen = sizeof(wheel) / sizeof(int);
  int next = 0;
  if (n == 1) {
    return 0;
  }
  if (n == 2 || n == 3) {
    return n;
  }

  while (factor <= n) {
    while (n % factor == 0) {
      biggest = factor;
      n /= factor;
    }
    factor += (unsigned long) wheel[next];
    next++;
    if (next == wlen) {
      next = start;
    }
  }
  if (n > 1 && biggest == 0) {
    return n;
  }

  return biggest;
}



int isprime(unsigned long n)
{
  unsigned long i;
  unsigned long isqrt = (unsigned long) floor(sqrt(n)) + 1;

  for (i = 2; i < isqrt; i++) {
    if ((n % i) == 0) {
      return 0;
    }
  }
  return 1;
}

#include <time.h>
int main()
{
  unsigned long n, i, lpf = 0, wheelfactor = 0;
  clock_t start,stop;

  if(scanf("%lu", &n) != 1){
     fprintf(stderr,"Must be a positive, small integer \n");
     exit(EXIT_FAILURE);
  }

  start = clock();
  wheelfactor = primefactors(n);
  stop = clock();
  printf("WHEEL Time %.10f seconds\n", (double) (stop - start) / CLOCKS_PER_SEC);
  printf("WHEEL %lu\n",wheelfactor );

  start = clock();
  if (n == 1) {
    puts("0");
    goto END;
  }
  if (n == 2 || ISPRIME(n) == 1) {
    printf("NAIVE %lu (n is prime, next print must show 0)\n", n);
    goto END;
  }
  for (i = 2; i < n / 2 + 1; i++) {
    if (ISPRIME(i) == 1 && n % i == 0) {
      //printf("PRIME %lu\n", i);
      lpf = i;
      n /= i;
      // n might be down to 2 and isprime(2) returns true
      // but  only odd primes are allowed at this point
      if (ISPRIME(n) == 1 && n > lpf) {
        //printf("ISPRIME %lu\n", n);
        lpf = n;
      }
    }
  }
END:
  stop = clock();

  printf("NAIVE Time %.10f seconds\n", (double) (stop - start) / CLOCKS_PER_SEC);
  printf("NAIVE %lu\n", lpf);
  return 0;
}

您可以尝试使用一些大型复合材料，例如：11529215046068460 以查看运行时的差异，或者 1152921123 这有点极端，需要很大的耐心——可能需要几分钟。