Question

我编译了程序但是当我给出输入= 600851475143时没有结果。程序是找到最大的素因子（13195的素数因子是5,7,13和29.什么是最大的素数因子号码600851475143？）有什么问题？

#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，其运行次数大致与你输入的数字相同，i从2变为n-1。在for循环中，您调用一个函数，该函数大约运行i次，这与循环中的i相同。这意味着您的程序会进行n^2数学计算，其中n是您的输入。如果输入12位数字，n^2是24位数字的大小，对于程序，需要花费大量时间进行10^24计算。所以你的程序运行，它只是计算。

Answer 2

正如其他人已经说过的那样：你不需要走完整条路。只要达到n的平方根就足够了。每次找到因子时，您也可以减少#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; }的值。

稍微快一点的方法虽然和天真版本一样复杂，但是使用方向盘。这里使用的轮子是11个粗略数字之间的距离（加上前面的素数和下一轮的偏移量）。

所有在一起：

11529215046068460

您可以尝试使用一些大型复合材料，例如：1152921123来查看运行时间的差异，或者def function(): string = StringVar() string.set("Hello I'm A.py")更加极端且需要很多耐心 - 它可能需要一个几分钟。

Answer 3

早些时候，我注意到我的原始解决方案：

600851475143仍然需要太长时间，所以现在是时候寻找了完全不同的算法

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

这是一个简单快速的解决方案@ under5hell在该页面上的答案：

var myTable = $('#myTable').DataTable();

myTable.row( ':eq(0)' ).edit( {
    name: 'Edit first row name object'
} );

可以在几分之一秒内找到最大的素数因子600851475143：

<强>输出

#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;
}

在您查询SO之前，故事的道德似乎是搜索SO！

程序编译但不执行

3 个答案: