[POJ 1365] Prime Land
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 3211 | Accepted: 1473 |
Description
(ekx, ekx-1, ... ,e1, e0)
is considered to be the representation of x in prime base number system.
It is really true that all numerical calculations in prime base number system can seem to us a little bit unusual, or even hard. In fact, the children in Prime Land learn to add to subtract numbers several years. On the other hand, multiplication and division is very simple.
Recently, somebody has returned from a holiday in the Computer Land where small smart things called computers have been used. It has turned out that they could be used to make addition and subtraction in prime base number system much easier. It has been decided to make an experiment and let a computer to do the operation ``minus one''.
Help people in the Prime Land and write a corresponding program.
For practical reasons we will write here the prime base representation as a sequence of such pi and ei from the prime base representation above for which ei > 0. We will keep decreasing order with regard to pi.
Input
Output
Sample Input
17 1
5 1 2 1
509 1 59 1
0
Sample Output
2 4
3 2
13 1 11 1 7 1 5 1 3 1 2 1
Source
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define ll long long
#define N 100000 ll tot;
bool isprime[N+];
ll prime[N+]; //1~tot-1
void getprime() //复杂度:O(n)
{
tot=;
memset(isprime,true,sizeof(isprime));
isprime[]=isprime[]=false;
for(ll i=;i<=N;i++)
{
if(isprime[i]) prime[tot++]=i;
for(ll j=;j<tot;j++)
{
if(i*prime[j]>N) break;
isprime[i*prime[j]]=false;
if(i%prime[j]==)
{
break;
}
}
}
}
ll fatcnt;
ll factor[N][]; //0~fatcnt-1
ll getfactors(ll x) //x>1
{
fatcnt=;
ll tmp=x;
for(ll i=;prime[i]<=tmp/prime[i];i++)
{
factor[fatcnt][]=;
if(tmp%prime[i]==)
{
factor[fatcnt][]=prime[i];
while(tmp%prime[i]==)
{
factor[fatcnt][]++;
tmp/=prime[i];
}
fatcnt++;
}
}
if(tmp!=)
{
factor[fatcnt][]=tmp;
factor[fatcnt++][]=;
}
return fatcnt;
}
ll pow(ll a,ll b)
{
ll ret=;
while(b)
{
if(b&) ret*=a;
a=a*a;
b>>=;
}
return ret;
}
int main()
{
getprime();
ll num,a,b,i;
char op;
while(scanf("%lld",&a),a)
{
scanf("%lld%c",&b,&op);
num=pow(a,b);
if(op!='\n')
{
while(scanf("%lld%lld%c",&a,&b,&op))
{
num*=pow(a,b);
if(op=='\n') break;
}
}
getfactors(num-);
for(i=fatcnt-;i>;i--) printf("%lld %lld ",factor[i][],factor[i][]);
printf("%lld %lld\n",factor[i][],factor[i][]);
}
return ;
}
[POJ 1365] Prime Land的更多相关文章
- POJ 1365 Prime Land(数论)
题目链接: 传送门 Prime Land Time Limit: 1000MS Memory Limit: 10000K Description Everybody in the Prime ...
- POJ 1365 Prime Land(整数拆分)
题意:感觉题意不太好懂,题目并不难,就是给一些p和e,p是素数,e是指数,然后把这个数求出来,设为x,然后让我们逆过程输出x-1的素数拆分形式,形式与输入保持一致. 思路:素数打表以后正常拆分即可. ...
- 筛选法 || POJ 1356 Prime Land
英文题读不懂题==质数幂的形式给你一个数 把它减一再用质数幂的形式表示出来 *解法:质数从小到大模拟除一遍,输入有点别扭 #include <iostream> #include < ...
- [暑假集训--数论]poj1365 Prime Land
Everybody in the Prime Land is using a prime base number system. In this system, each positive integ ...
- 双向广搜 POJ 3126 Prime Path
POJ 3126 Prime Path Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16204 Accepted ...
- poj 2689 Prime Distance(大区间素数)
题目链接:poj 2689 Prime Distance 题意: 给你一个很大的区间(区间差不超过100w),让你找出这个区间的相邻最大和最小的两对素数 题解: 正向去找这个区间的素数会超时,我们考虑 ...
- POJ 3126 Prime Path(素数路径)
POJ 3126 Prime Path(素数路径) Time Limit: 1000MS Memory Limit: 65536K Description - 题目描述 The minister ...
- Miller_rabin算法+Pollard_rho算法 POJ 1811 Prime Test
POJ 1811 Prime Test Time Limit: 6000MS Memory Limit: 65536K Total Submissions: 32534 Accepted: 8 ...
- POJ 3518 Prime Gap(素数)
POJ 3518 Prime Gap(素数) id=3518">http://poj.org/problem? id=3518 题意: 给你一个数.假设该数是素数就输出0. 否则输出比 ...
随机推荐
- CorelDRAW 二维码插件
随着智能手机的流行,二维码在各个领域大量应用,这个插件在补CorelDRAW这方面的不足: 这个插件是 cpg 格式,安装请看这篇博客:http://www.cnblogs.com/o594cql/p ...
- Spark Streaming揭秘 Day26 JobGenerator源码图解
Spark Streaming揭秘 Day26 JobGenerator源码图解 今天主要解析一下JobGenerator,它相当于一个转换器,和机器学习的pipeline比较类似,因为最终运行在Sp ...
- 【C++】GacLib——ListView.ViewSwitching
http://www.gaclib.net/Demos/Controls.ListView.ViewSwitching/Demo.html#FILESYSTEMINFORMATION_H
- Spring execution 表达式
execution(modifiers-pattern? ret-type-pattern declaring-type-pattern? name-pattern(param-pattern) th ...
- sql with递归
with temp as ( select Id, UserId, OfficeID, RoleId, DeptId, IsDelete, IsEnd, ParentId from [dbo]. ...
- webstorm 11 安装配置 grunt 时遇到的问题及解决办法
想学grunt的可以看看这篇文章,写的很有意思,教程之类的我就不写了,网上很多资料,我就记录下我遇到的问题和解决办法. http://yujiangshui.com/grunt-basic-tutor ...
- protobuf 向前兼容向后兼容
http://blog.163.com/jiang_tao_2010/blog/static/12112689020114305013458/ 不错的protobuf.. protobuf的编码方式: ...
- java.lang.NoSuchMethodError: javaxservlet.http.HttpServletRequest.isAsyncStarted()Z
鸣谢网址:http://stackoverflow.com/questions/25940571/java-lang-nosuchmethoderror-javaxservlet-http-https ...
- Hibernate一对一单向外键关联
一.一对一单向外键关联: 一对一单向外键关联主要用到了以下两个注解: 1.OneToOne(cascade=CasecadeTYPE.ALL); cascade=CasecadeTYPE.ALL:表示 ...
- win8连接蓝牙听歌
今天买了一个蓝牙耳机,琢磨着在win8.1上听一下,可是折腾了一阵时间,现在把最佳配置方式写出来,希望对朋友有所帮助 确保win8的蓝牙驱动已经安装完毕,并且开启蓝牙,win8,设置--右下角更改电脑 ...