hdu 3199 Hamming Problem(构造?枚举?)
2024-09-08 17:10:33
题意:
For each three prime numbers p1, p2 and p3, let's define Hamming sequence Hi(p1, p2, p3), i=1, ... as containing in increasing order all the natural numbers whose only prime divisors are p1, p2 or p3.
For example, H(2, 3, 5) = 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, ...
So H5(2, 3, 5)=6.
给p1,p2,p3,n,求Hn(p1,p2,p3)
思路:
和humble numbers那题一样,,,看代码,,
代码:
int p1,p2,p3,x;
ll a[1000005]; int main(){ while(scanf("%d%d%d%d",&p1,&p2,&p3,&x)!=EOF){
if(p1>p2) swap(p1,p2);
if(p1>p3) swap(p1,p3);
if(p2>p3) swap(p2,p3);
a[1]=1;
int cn=1;
while(cn<=x){
ll temp;
ll ans=INF;
rep2(i,cn,1){
temp=a[i]*p1; if(temp>a[cn]) ans=min(ans,temp);
temp=a[i]*p2; if(temp>a[cn]) ans=min(ans,temp);
temp=a[i]*p3; if(temp>a[cn]) ans=min(ans,temp);
if(temp<=a[cn]) break;
}
a[++cn]=ans;
}
printf("%I64d\n",a[cn]);
} return 0;
}
最新文章
- 如何查看 Linux是32位还是64位?
- apache端口的修改
- 基于JQuery实现滚动到页面底端时自动加载更多信息
- Linux USB摄像头驱动【转】
- 视频相关android软件
- 【Populating Next Right Pointers in Each Node II】cpp
- Knockout.js 初探
- ThinkPHP3.2.3新特性之:数据库设置
- 只允许输入数字的TextBox控件
- XAMPP的Apache服务器无法正常启动解决方案
- Java版冒泡排序和选择排序
- Unity3d 物体沿着正七边形轨迹移动
- WebApi的版本控制
- Dubbo2.6.5入门——简单的HelloWorld
- css冲刺
- 安装Logtail(Linux系统)
- LeetCode 112 Minimum Depth of Binary Tree
- Alpha 冲刺报告(2/10)
- php 数组与数组 的交集和差集
- 一篇搞定SQLAlchemy--关系对象映射