zoukankan      html  css  js  c++  java
  • 1059 Prime Factors (25 分)素数 “筛选法”

    1059 Prime Factors (25 分)

    Given any positive integer NNN, you are supposed to find all of its prime factors, and write them in the format NNN = p1k1×p2k2×⋯×pmkm{p_1}^{k_1} imes {p_2}^{k_2} imes cdots imes {p_m}^{k_m}p1​​k1​​​​×p2​​k2​​​​××pm​​km​​​​.

    Input Specification:

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

    Output Specification:

    Factor NNN in the format NNN = p1p_1p1​​^k1k_1k1​​*p2p_2p2​​^k2k_2k2​​**pmp_mpm​​^kmk_mkm​​, where pip_ipi​​'s are prime factors of NNN in increasing order, and the exponent kik_iki​​ is the number of pip_ipi​​ -- hence when there is only one pip_ipi​​, kik_iki​​ is 1 and must NOT be printed out.

    Sample Input:

    97532468
    

    Sample Output:

    97532468=2^2*11*17*101*1291
    思路:
      1不是素数
    #include<iostream>
    #include<vector>
    #include<algorithm>
    #include<map>
    #include<set>
    #include<cmath>
    #include<climits>
    #include<sstream>
    #include<cstdio>
    #include<string.h>
    #include<unordered_map>
    using namespace std;
    unordered_map<string,set<int>>mp;
    int main()
    {
        long long int n;
        scanf("%lld",&n);
        int num=sqrt(n);
        vector<int> primes(num,1);
        for(int i=2;i<=sqrt(num);i++)
        {
            for(int j=i+i;j<=sqrt(num);j+=i)
                primes[j]=0;
        }
        if(n==1)
            printf("1=1");
        else
        {
            printf("%lld=",n);
            int state=0;
            for(int i=2;i<=num;i++)
            {
                int cnt=0;
                bool flag=false;
                while(primes[i]==1&&n%i==0)
                {
                    n/=i;
                    cnt++;
                    flag=true;
                }
                if(flag)
                {
                    if(state!=0) printf("*");
                    state=1;
                    printf("%d",i);
                    if(cnt>1)
                        printf("^%d",cnt);
                }
                if(n==1)
                    break;
            }
            if(n!=1)
            {
                if(state!=0) printf("*");
                printf("%d",n);
            }
        }
        return 0;
    }
     
  • 相关阅读:
    10、驱动中的阻塞与非阻塞IO
    8、Linux设备驱动的并发控制
    入职一个月考核学习
    5、映射的思考
    6、udev机制
    7、字符设备系统
    linux 内存管理之kmalloc、vmalloc、malloc、get_gree_pages的区别
    嵌入式笔试题(linux基础)
    驱动总结
    系统移植总结
  • 原文地址:https://www.cnblogs.com/zhanghaijie/p/10346326.html
Copyright © 2011-2022 走看看