PTA(Advanced Level)1059.Prime Factors

标签：cas   file   level   osi   std   while   href   problems   质因数分解   

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×?×*pmk**m*.

Input Specification:

Each input file contains one test case which gives a positive integer N in the range of long int.

Output Specification:

Factor N in the format N = p1^k1*p2^k2*…*pm^km, where *p**i‘s are prime factors of N* in increasing order, and the exponent *k**i* is the number of *pi* -- hence when there is only one pi, ki is 1 and must NOT** be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291

思路

• 质因数分解和因数分解类似，只是要先得到质数表，我这里使用的是质数筛法
• \(N\)的最大是\(2^{31}-1\)，\(\sqrt{n}\approx2^{15.5}\)，除非这个数本身是质数，不然我们因子判断到\(\sqrt{n}\)就好了

代码

#include<bits/stdc++.h>
using namespace std;
struct factor
{
    int x;  //质数
    int n;  //阶数
}fac[50];   
int primes[100100];
int prime_length = 0;
void gen_prime()
{
    int MAXN = 100100;
    bool flag[MAXN] = {0};
    for(int i=2;i<MAXN;i++)
    {
        if(!flag[i])
        {
            primes[prime_length++] = i;
            for(int j=i+i;j<MAXN;j+=i)  flag[j] = true;
        }
    }
}//质数筛法

int main()
{
    int n, n_copy;
    cin >> n;
    n_copy = n;
    gen_prime();

    int last_pos = 0;
    if(n == 1)  cout << "1=1";      //1是特殊情况
    else
    {
        int sqrt_value = (int)sqrt(n * 1.0);    //一个数的因子除了自身不会超过自身的根号n
        for(int i=0;primes[i]<sqrt_value && i<prime_length;i++)     //不超过根号n，以及不超过质数表的长度
        {
            while(n % primes[i] == 0)
            {
                fac[last_pos].x = primes[i];
                fac[last_pos].n == 0;
                while(n % primes[i] == 0)
                {
                    fac[last_pos].n++;
                    n /= primes[i];
                }
                last_pos++;
            }
            if(n == 1)  break;
        }

        if(n != 1)  //说明自己本身就是质数了
        {
            fac[last_pos].x = n;
            fac[last_pos].n = 1;
            last_pos++;
        }


        cout << n_copy << "=";
        for(int i=0;i<last_pos;i++)
        {
            if(fac[i].n != 1)
                printf("%d^%d", fac[i].x, fac[i].n);
            else
                printf("%d", fac[i].x);
            if(i != last_pos - 1)
                printf("*");
        }
    }
    return 0;
}

引用

https://pintia.cn/problem-sets/994805342720868352/problems/994805415005503488

PTA(Advanced Level)1059.Prime Factors

标签：cas   file   level   osi   std   while   href   problems   质因数分解   

原文地址：https://www.cnblogs.com/MartinLwx/p/12538484.html