1091. Tmutarakan Exams
1091. Tmutarakan ExamsTime limit: 1.0 second
Memory limit: 64 MB University of New Tmutarakan trains the first-class specialists in mental arithmetic. To enter the University you should master arithmetic perfectly. One of the entrance exams at the Divisibility Department is the following. Examinees are asked to find K different numbers that have a common divisor greater than 1. All numbers in each set should not exceed a given number S. The numbers K and S are announced at the beginning of the exam. To exclude copying (the Department is the most prestigious in the town!) each set of numbers is credited only once (to the person who submitted it first).
Last year these numbers were K=25 and S=49 and, unfortunately, nobody passed the exam. Moreover, it was proved later by the best minds of the Department that there do not exist sets of numbers with the required properties. To avoid embarrassment this year, the dean asked for your help. You should find the number of sets of K different numbers, each of the numbers not exceeding S, which have a common divisor greater than 1. Of course, the number of such sets equals the maximal possible number of new students of the Department.
InputThe input contains numbers K and S (2 ≤ K ≤ S ≤ 50).
OutputYou should output the maximal possible number of the Department's new students if this number does not exceed 10000 which is the maximal capacity of the Department, otherwise you should output 10000.
Sample
Problem Author: Stanislav Vasilyev
Problem Source: USU Open Collegiate Programming Contest March'2001 Senior Session |
2000–2016 Timus Online Judge Team. All rights reserved.
思路:和http://www.cnblogs.com/zzuli2sjy/p/5467008.html一样;
1 #include<stdio.h>
2 #include<string.h>
3 #include<iostream>
4 #include<algorithm>
5 #include<queue>
6 using namespace std;
7 typedef long long LL;
8 bool prime[100];
9 int ans[100];
10 int coutt[10000];
11 LL dp[60][60];
12 int ask[100];
13 int id[100];
14 queue<int>que;
15 int main(void)
16 {
17 int i,j,k,p,q;
18 dp[0][0]=1;
19 dp[1][0]=1;
20 dp[1][1]=1;
21 for(i=2; i<=60; i++)
22 {
23 for(j=0; j<=60; j++)
24 {
25 if(j==0||i==j)
26 {
27 dp[i][j]=1;
28 }
29 else dp[i][j]=dp[i-1][j-1]+dp[i-1][j];
30 }
31 }
32 for(i=2; i<=10; i++)
33 {
34 if(!prime[i])
35 {
36 for(j=i; i*j<=50; j++)
37 {
38 prime[i*j]=true;
39 }
40 }
41 }
42 int cnt=0;
43 for(i=2; i<=50; i++)
44 {
45 if(!prime[i])
46 {
47 ans[cnt++]=i;
48 }
49 }
50 while(scanf("%d %d",&p,&q)!=EOF)
51 { int s;
52 memset(coutt,0,sizeof(coutt));
53 for(s=2; s<=q; s++)
54 {
55 int cc=s;
56 int flag=0;
57 int t=0;
58 while(cc>1)
59 {
60 if(cc%ans[t]==0&&flag==0)
61 {
62 flag=1;
63 que.push(ans[t]);
64 cc/=ans[t];
65 }
66 else if(cc%ans[t]==0)
67 {
68 cc/=ans[t];
69 }
70 else
71 {
72 t++;
73 flag=0;
74 }
75 }
76 int vv=0;
77 while(!que.empty())
78 {
79 ask[vv++]=que.front();
80 que.pop();
81 }
82 for(i=1; i<=(1<<vv)-1; i++)
83 {
84 LL sum=1;
85 int dd=0;
86 for(j=0; j<vv; j++)
87 {
88 if(i&(1<<j))
89 {
90 dd++;
91 sum*=ask[j];
92 }
93 }
94 id[sum]=dd;
95 coutt[sum]++;
96
97 }
98 }
99 LL summ=0;
100 for(i=2; i<=50; i++)
101 {
102 if(id[i]%2&&coutt[i]>=p)
103 {
104 summ+=dp[coutt[i]][p];
105 }
106 else if(coutt[i]>=p)summ-=dp[coutt[i]][p];
107 }if(summ>=10000)summ=10000;
108 printf("%lld\n",summ);
109 }
110 return 0;
111 }
1091. Tmutarakan Exams的更多相关文章
- ural 1091. Tmutarakan Exams(容斥原理)
1091. Tmutarakan Exams Time limit: 1.0 secondMemory limit: 64 MB University of New Tmutarakan trains ...
- Ural 1091 Tmutarakan Exams
Tmutarakan Exams Time Limit: 1000ms Memory Limit: 16384KB This problem will be judged on Ural. Origi ...
- ural 1091. Tmutarakan Exams 和 codeforces 295 B. Greg and Graph
ural 1091 题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1091 题意是从1到n的集合里选出k个数,使得这些数满足gcd大于1 ...
- ural 1091. Tmutarakan Exams(容斥)
http://acm.timus.ru/problem.aspx? space=1&num=1091 从1~s中选出k个数,使得k个数的最大公约数大于1,问这种取法有多少种. (2<=k ...
- URAL - 1091 Tmutarakan Exams (简单容斥原理)
题意:K个不同数组成的集合,每个数都不超过S且它们的gcd>1.求这样的数的个数 分析:从2开始枚举gcd,但这样会发生重复.譬如,枚举gcd=2的集合个数和gcd=3的集合个数,枚举6的时候就 ...
- F - Tmutarakan Exams URAL - 1091 -莫比乌斯函数-容斥 or DP计数
F - Tmutarakan Exams 题意 : 从 < = S 的 数 中 选 出 K 个 不 同 的 数 并 且 gcd > 1 .求方案数. 思路 :记 录 一 下 每 个 数 的 ...
- Tmutarakan Exams URAL - 1091(莫比乌斯函数 || 容斥)
题意: 求1 - s 中 找出k个数 使它们的gcd > 1 求这样的k个数的对数 解析: 从每个素数的倍数中取k个数 求方案数 然后素数组合,容斥一下重的 奇加偶减 莫比乌斯函数的直接套模 ...
- 2014 Super Training #3 H Tmutarakan Exams --容斥原理
原题: URAL 1091 http://acm.timus.ru/problem.aspx?space=1&num=1091 题意:要求找出K个不同的数字使他们有一个大于1的公约数,且所有 ...
- URAL1091. Tmutarakan Exams(容斥)
1091 容斥原理 #include <iostream> #include<cstdio> #include<cstring> #include<algor ...
随机推荐
- Visual Studio Code常用操作整理
Live Server插件可以在保存html文件后实时地刷新页面 在html文件中键入"! +Tap"会生成一个html模板 保存文件:Ctrl+S 文件跳转:Ctrl+P 文件内 ...
- 表格合并单元格【c#】
gridBranchInfo.DataSource = dtBranchViewList; gridBranchInfo.DataBind(); Random random = new Random( ...
- Hadoop入门 运行环境搭建
模板虚拟机 目录 模板虚拟机 1 硬件 2 操作系统 3 IP地址和主机名称 vm windows10 Hadoop100服务器 远程访问工具 其他准备 克隆虚拟机 克隆 修改主机名/ip 安装jdk ...
- 16. Linux find查找文件及文件夹命令
find的主要用来查找文件,查找文件的用法我们比较熟悉,也可用它来查找文件夹,用法跟查找文件类似,只要在最后面指明查找的文件类型 -type d,如果不指定type类型,会将包含查找内容的文件和文件夹 ...
- android:textAppearance解析
Android的系统自带的文字外观设置及实际显示效果图 android:textAppearancexml布局里面设置文字的外观: 如"android:textAppearance=&quo ...
- gitlab之实战部署
#:准备Java环境,安装jdk root@ubuntu:~# cd /usr/local/src/ root@ubuntu:/usr/local/src# ls jdk-8u191-linux-x6 ...
- 监控网站是否异常的shell脚本
本节内容:shell脚本监控网站是否异常,如有异常就自动发邮件通知管理员. 脚本检测流程,如下:1,检查网站返回的http_code是否等于200,如不是200视为异常.2,检查网站的访问时间,超过M ...
- 【Service】【Database】【MySQL】基础概念
1. 数据模型:层次模型.网状模型.关系模型 关系模型: 二维关系: 表:row, column 索引:index 视图:view 2. SQL接口:Structured Query Language ...
- jstl中的foreach标签
<%@ page import="java.util.ArrayList" %><%@ page import="java.util.List" ...
- 【C/C++】习题3-5 谜题/算法竞赛入门经典/数组和字符串
[题目] 有一个5*5的网络,恰好有一个格子是空的(空格),其他格子各有一个字母. 指令:A, B, L, R 把空格上.下.左.右的相邻字母移到空格中. [输入] 初始网格和指令序列(以数字0结束) ...