zoukankan      html  css  js  c++  java
  • hdu 5428 The Factor(数学)

    Problem Description
    There is a sequence of n positive integers. Fancycoder is addicted to learn their product, but this product may be extremely huge! However, it is lucky that FancyCoder only needs to find out one factor of this huge product: the smallest factor that contains more than 2 factors(including itself; i.e. 4 has 3 factors so that it is a qualified factor). You need to find it out and print it. As we know, there may be none of such factors; in this occasion, please print -1 instead. 
     
    Input
    The first line contains one integer T (1T15), which represents the number of testcases. 
    For each testcase, there are two lines:
    1. The first line contains one integer denoting the value of n (1n100).
    2. The second line contains n integers a1,,an (1a1,,an2×109), which denote these n positive integers. 
     
    Output
    Print T answers in T lines.
     
    Sample Input
    2 3 1 2 3 5 6 6 6 6 6
     
    Sample Output
    6 4
     
    Source
     

    对于每一个数字,它有用的部分其实只有它的所有质因子(包括相等的)。求出所有数的所有质因子中最小的两个,相乘就是答案。如果所有数字的质因子个数不到两个,那么就是无解。时间复杂度

    O(n*sqrt(a))O(nsqrt(a))

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<set>
     5 #include<cmath>
     6 #include<algorithm>
     7 using namespace std;
     8 #define ll long long 
     9 multiset<ll> s;
    10 multiset<ll>::iterator it,it1;
    11 int main()
    12 {
    13     int t;
    14     scanf("%d",&t);
    15     while(t--){
    16         s.clear();
    17         ll n;
    18         scanf("%I64d",&n);
    19         for(ll i=0;i<n;i++){
    20             ll m;
    21             scanf("%I64d",&m);
    22             for(ll j=2;j*j<=m;j++){
    23                 if(m%j==0){
    24                     while(m%j==0){
    25                         s.insert(j);
    26                         m/=j;
    27                     }
    28                 }
    29             }
    30             if(m>1) s.insert(m);
    31         }
    32         ll size=s.size();
    33         
    34         if(size<2) printf("-1
    ");
    35         else{
    36             it=s.begin();
    37             ll ans1=(*it);
    38             it++;
    39             ll ans2=(*it);
    40             printf("%I64d
    ",ans1*ans2);
    41         } 
    42     }
    43     return 0;
    44 }
    View Code
  • 相关阅读:
    在sql中日期转化
    SQL Server标准版企业版个人版区别
    Oracle 9i默认用户
    SQL Server和Oracle的常用函数对比
    Delphi6函数大全
    Oracle 不支持 TOP 关键字
    c#连接数据库
    VC6安装及打开工程文件问题的解决
    error C4430: 缺少类型说明符 假定为 int。注意: C++ 不支持默认 int
    VC++ dll导出类
  • 原文地址:https://www.cnblogs.com/UniqueColor/p/4784042.html
Copyright © 2011-2022 走看看