2017ACM/ICPC广西邀请赛-重现赛(感谢广西大学)
上一场CF打到心态爆炸,这几天也没啥想干的
A Math Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Each case only contains a positivse integer n in a line.
1≤n≤1018
4
2
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=;
int main()
{
ll n;
while(~scanf("%lld",&n))
{
if(n>=N)
printf("15\n");
else
{
for(int k=; k<; k++)
{
ll s=;
for(int i=; i<k; i++)
s*=k;
if(s>n)
{
printf("%d\n",k-);
break;
}
}
}
}
return ;
}
Covering
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
To protect boys and girls from getting hurt when playing happily on the playground, rich boy Bob decided to cover the playground using his carpets.
Meanwhile, Bob is a mean boy, so he acquired that his carpets can not overlap one cell twice or more.
He has infinite carpets with sizes of 1×2 and 2×1, and the size of the playground is 4×n.
Can you tell Bob the total number of schemes where the carpets can cover the playground completely without overlapping?
Each test case only contains one positive integer n in a line.
1≤n≤1018
2
5
a[n]=a[n-1]+5*a[n-2]+a[n-3]-a[n-4];
#include <stdio.h>
#include <string.h>
const int MD=1e9+;
typedef long long LL;
struct matrix
{
LL mat[][];
};
matrix matmul(matrix a,matrix b,int n)
{
int i,j,k;
matrix c;
memset(c.mat,,sizeof(c.mat));
for(i=; i<n; i++)
{
for(j=; j<n; j++)
{
for(k=; k<n; k++)
{
c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%MD;
}
}
}
return c;
}
matrix matpow(matrix a,LL k,int n)
{
matrix b;
int i;
memset(b.mat,,sizeof(b.mat));
for(i=; i<n; i++) b.mat[i][i]=;
while(k)
{
if(k&) b=matmul(a,b,n);
a=matmul(a,a,n);
k>>=;
}
return b;
}
int main()
{
LL k;
matrix a,b;
memset(a.mat,,sizeof(a.mat));
memset(b.mat,,sizeof(b.mat));
a.mat[][]=,a.mat[][]=,a.mat[][]=;
b.mat[][]=,b.mat[][]=,b.mat[][]=,b.mat[][]=-;
b.mat[][]=,b.mat[][]=,b.mat[][]=;
while(~scanf("%lld",&k))
{
printf("%lld\n",(matmul(matpow(b,k,),a,).mat[][]+MD)%MD);
}
return ;
}
CS Course
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Today he has learned bit-operations in Algorithm Lessons, and he got a problem as homework.
Here is the problem:
You are giving n non-negative integers a1,a2,⋯,an, and some queries.
A query only contains a positive integer p, which means you
are asked to answer the result of bit-operations (and, or, xor) of all the integers except ap.
Each test case begins with two positive integers n and p
in a line, indicate the number of positive integers and the number of queries.
2≤n,q≤105
Then n non-negative integers a1,a2,⋯,an follows in a line, 0≤ai≤109 for each i in range[1,n].
After that there are q positive integers p1,p2,⋯,pqin q lines, 1≤pi≤n for each i in range[1,q].
1 1 1
1
2
3
1 1 0
1 1 0
异或最简单,再异或一次就好了
所以按位存储了
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+;
int a[N],b[N];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
memset(b,,sizeof(b));
int Xor=,And=0xffffffff,Or=;
for(int i=; i<=n; i++)
{
int x;
scanf("%d",&x);
a[i]=x;
And&=x;
Or|=x;
Xor^=x;
for(int j=; x; j++,x>>=)
b[j]+=x%;
}
while(m--)
{
int q;
scanf("%d",&q);
q=a[q];
int A=And,O=Or,X=Xor;
X=X^q;
for(int j=; j<=; j++,q>>=)
{
if(b[j]==n-&&q%==)A+=(<<j);
if(b[j]==&&q%)O-=(<<j);
}
printf("%d %d %d\n",A,O,X);
}
}
return
Duizi and Shunzi
Now give you n integers, ai(1≤i≤n)ai(1≤i≤n)
We define two identical numbers (eg: 2,22,2) a Duizi,
and three consecutive positive integers (eg: 2,3,42,3,4) a Shunzi.
Now you want to use these integers to form Shunzi and Duizi as many as possible.
Let s be the total number of the Shunzi and the Duizi you formed.
Try to calculate max(s)max(s).
Each number can be used only once.
InputThe input contains several test cases.
For each test case, the first line contains one integer n(1≤n≤1061≤n≤106).
Then the next line contains n space-separated integers aiai (1≤ai≤n1≤ai≤n)
OutputFor each test case, output the answer in a line.
Sample Input
7
1 2 3 4 5 6 7
9
1 1 1 2 2 2 3 3 3
6
2 2 3 3 3 3
6
1 2 3 3 4 5
Sample Output
2
4
3
2
Hint
Case 1(1,2,3)(4,5,6) Case 2(1,2,3)(1,1)(2,2)(3,3) Case 3(2,2)(3,3)(3,3) Case 4(1,2,3)(3,4,5)
这个题看起来很简单,问你最多可形成多少个对子和顺子
可是有坑啊,可以按照对子打,也可以按照顺子打,我当然按照对子打了,但是按照对子打可能我的顺子就没了,所以我首先是要打足够多的牌
比如我往下贪心的时候,如果第二张恰好是对子,我贪心就亏了,但是我下一张正好三张我肯定就要了这个顺子
所以就是记录顺子和找对子了
#include <stdio.h>
#include <string.h>
const int N=1e5+;
int a[N];
int main()
{
int n;
while(~scanf("%d",&n))
{
memset(a,,sizeof(int)*(n+));
for(int i=; i<n; i++)
{
int x;
scanf("%d",&x);
a[x]++;
}
int ans=;
for(int i=; i<n-; i++)
{
ans+=a[i]/;
if(a[i]&&&a[i+]&&&a[i+])
{
ans++;
a[i+]--;
a[i+]--;
}
}
ans+=a[n-]/+a[n]/;
printf("%d\n",ans);
}
return ;
}
2017ACM/ICPC广西邀请赛-重现赛(感谢广西大学)的更多相关文章
- 2017ACM/ICPC广西邀请赛-重现赛 1007.Duizi and Shunzi
Problem Description Nike likes playing cards and makes a problem of it. Now give you n integers, ai( ...
- 2017ACM/ICPC广西邀请赛-重现赛 1010.Query on A Tree
Problem Description Monkey A lives on a tree, he always plays on this tree. One day, monkey A learne ...
- 2017ACM/ICPC广西邀请赛-重现赛 1004.Covering
Problem Description Bob's school has a big playground, boys and girls always play games here after s ...
- 2017ACM/ICPC广西邀请赛-重现赛
HDU 6188 Duizi and Shunzi 链接:http://acm.hdu.edu.cn/showproblem.php?pid=6188 思路: 签到题,以前写的. 实现代码: #inc ...
- 2017ACM/ICPC广西邀请赛-重现赛 1001 A Math Problem
2017-08-31 16:48:00 writer:pprp 这个题比较容易,我用的是快速幂 写了一次就过了 题目如下: A Math Problem Time Limit: 2000/1000 M ...
- 2017ACM/ICPC广西邀请赛-重现赛1005 CS course
2017-08-31 16:19:30 writer:pprp 这道题快要卡死我了,队友已经告诉我思路了,但是做题速度很缓慢,很费力,想必是因为之前 的训练都是面向题解编程的缘故吧,以后不能这样了,另 ...
- HDU 6191 2017ACM/ICPC广西邀请赛 J Query on A Tree 可持久化01字典树+dfs序
题意 给一颗\(n\)个节点的带点权的树,以\(1\)为根节点,\(q\)次询问,每次询问给出2个数\(u\),\(x\),求\(u\)的子树中的点上的值与\(x\)异或的值最大为多少 分析 先dfs ...
- 2017ACM/ICPC广西邀请赛
A.A Math Problem #include <bits/stdc++.h> using namespace std; typedef long long ll; inline ll ...
- 2017ACM/ICPC广西邀请赛 Duizi and Shunzi
题意:就是一个集合分开,有两种区分 对子:两个相同数字,顺子:连续三个不同数字,问最多分多少个 解法:贪心,如果当前数字不构成顺子就取对子 /2,如果可以取顺子,那么先取顺子再取对子 #include ...
随机推荐
- 服务器 未能加载文件或程序集“XXXX”或它的某一个依赖项。试图加载格式不正确的程序。
,本人采用的第一种解决办法解决,已解决 问题2: 在同一个服务器上想要一个IP有两个网址,配置端口号,给新端口号开权限
- Java运算符——通过示例学习Java编程(6)
作者:CHAITANYA SINGH 来源:https://www.koofun.com/pro/kfpostsdetail?kfpostsid=17 运算符是表示动作的字符,例如+是表示加法的算 ...
- Android Theme.Dialog 到光 AppCompatDialog
我用在我的 style.xml 作为主要应用程序主题 <style name="AppTheme" parent="Theme.AppCompat.Light&qu ...
- 分析(ExtractTransformLoad)与挖掘(DataMine)有何区别 ?
首先,介绍一下ETL 和 DM: ETL/Extraction-Transformation-Loading——用于完成DB到DW的数据转存,它将DB中的某一个时间点的状态,“抽取”出来,根据 ...
- DataModel doesn't have preference values
mahout和hadoop实现简单的智能推荐系统的时候,出现了一下几个方面的错误 DataModel doesn't have preference values 意思是DataModel中没有找到初 ...
- spring 中bean学习笔记
spring 中bean 一.bean的定义和应用 1. bean 形象上类似于getXX()和setXX()的一种. 2. 由于java是面向对象的,类的方法和属性在使用中需要实例化. 3. 规律: ...
- Dance links算法
其实Dance links只是一种数据结构,Dance links 才是一种算法.dacing links x就是一个高效的求解该类问题的算法,而这种算法,基于交叉十字循环双向 链表.下面是双向十字链 ...
- urlrrtrieve()实例_下载微博短视频
1.确定目标 在微博页面找一想要下载的短视频,通过审查元素找到视频的url. 如://f.us.sinaimg.cn/00150tBNlx07l0qjoSJi01040201m7z90k010.mp4 ...
- urllib基础-请求对象request
简单的案例-爬取百度首页 from urllib import request ''' 爬取百度首页 ''' # 确定爬去目标 base_url = 'http://www.baidu.com' # ...
- core 下使用 autofac
依赖注入小伙伴们比较常了,这里只说core 下autofac依赖注入的使用 ,不多费话,直接代码. 在 Startup.cs里 public void ConfigureServices(IServi ...