PAT (Advanced Level) 1059. Prime Factors (25)
素因子分解。
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<map>
#include<queue>
#include<string>
#include<stack>
#include<vector>
using namespace std; long long n; bool f(long long a)
{
for(long long i=;i*i<=a;i++)
if(a%i==) return ;
return ;
} struct X
{
long long num;
int tot;
}s[]; int main()
{
scanf("%lld",&n); int g=;
printf("%lld=",n);
for(long long i=;;i++)
{
if(f(n))
{
s[g].num=n;
s[g].tot=;
g++;
break;
}
if(n%i!=) continue;
if(f(i)==) continue;
s[g].num=i; s[g].tot=;
while()
{
if(n%i==)
{
n=n/i;
s[g].tot++;
}
else break;
}
g++;
if(n==) break;
} printf("%lld",s[].num);
if(s[].tot>) printf("^%d",s[].tot);
for(int i=;i<g;i++)
{
printf("*");
printf("%lld",s[i].num);
if(s[i].tot>) printf("^%d",s[i].tot);
}
printf("\n"); return ;
}
PAT (Advanced Level) 1059. Prime Factors (25)的更多相关文章
- 【PAT甲级】1059 Prime Factors (25 分)
题意: 输入一个正整数N(范围为long int),输出它等于哪些质数的乘积. trick: 如果N为1,直接输出1即可,数据点3存在这样的数据. 如果N本身是一个质数,直接输出它等于自己即可,数据点 ...
- PAT 甲级 1059 Prime Factors (25 分) ((新学)快速质因数分解,注意1=1)
1059 Prime Factors (25 分) Given any positive integer N, you are supposed to find all of its prime ...
- 1059 Prime Factors (25分)
1059 Prime Factors (25分) 1. 题目 2. 思路 先求解出int范围内的所有素数,把输入x分别对素数表中素数取余,判断是否为0,如果为0继续除该素数知道余数不是0,遍历到sqr ...
- PAT Advanced 1059 Prime Factors (25) [素数表的建⽴]
题目 Given any positive integer N, you are supposed to find all of its prime factors, and write them i ...
- PAT 1059. Prime Factors (25) 质因子分解
题目链接 http://www.patest.cn/contests/pat-a-practise/1059 Given any positive integer N, you are suppose ...
- 1059. Prime Factors (25)
时间限制 50 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 HE, Qinming Given any positive integer N, y ...
- PAT甲题题解-1059. Prime Factors (25)-素数筛选法
用素数筛选法即可. 范围long int,其实大小范围和int一样,一开始以为是指long long,想这就麻烦了该怎么弄. 而现在其实就是int的范围,那难度档次就不一样了,瞬间变成水题一枚,因为i ...
- PAT (Advanced Level) 1096. Consecutive Factors (20)
如果是素数直接输出1与素数,否则枚举长度和起始数即可. #include<cstdio> #include<cstring> #include<cmath> #in ...
- PAT (Advanced Level) 1114. Family Property (25)
简单DFS. #include<cstdio> #include<cstring> #include<cmath> #include<vector> # ...
随机推荐
- Chapter 2 Open Book——14
I backpedaled. "They seemed nice enough to me. I just noticed they keptto themselves. 我改口说道,他们看 ...
- Chapter 2 Open Book——11
"Hey, Dad, welcome home." hey爸爸,欢迎回家. "Thanks." He hung up his gun belt and step ...
- hdu_5695_Gym Class(拓扑排序)
题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=5695 题意:中文题,不解释 题解:逆向拓扑字典序就行 #include<cstdio> # ...
- 再谈KMP
昨天讲解了字典树和AC自动机后感觉整个人都蒙掉了.还好就是自己今天在网上看见一篇对KMP讲解非常详细的帖子,果断收藏.(点击这里查看) 然后代码的实现也就简单分析一些了,具体的知识点大家直接自己链接过 ...
- json 帮助工具
import java.lang.reflect.Type; import com.google.gson.Gson; /** * json 帮助工具 */public final class Gso ...
- sscanf用法
sscanf与scanf类似,都是用于输入的,只是后者以键盘(stdin)为输入源,前者以固定字符串为输入源. 1. 常见用法. 1 2 3 char buf[512] ; sscanf(" ...
- jquery datatable 参数api
jQuery 的插件 dataTables 是一个优秀的表格插件,提供了针对表格的排序.浏览器分页.服务器分页.筛选.格式化等功能.dataTables 的网站上也提供了大量的演示和详细的文档进行说明 ...
- linkButton
<?xml version="1.0" encoding="utf-8"?> <s:Application xmlns:fx="ht ...
- auto ash v1
startdate=$1enddate=$2#reporttype=$3#reportformat='text'oraclehome=`echo $ORACLE_HOME` dbid=`sqlplus ...
- regress
#! /bin/ksh ############### ### UAT ### ############### export ENVS=/test/change/env/env_test.sq ...