Codeforces Beta Round #46 (Div. 2)
Codeforces Beta Round #46 (Div. 2)
http://codeforces.com/contest/49
A
#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define maxn 1000005
typedef long long ll;
typedef unsigned long long ull; int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
map<char,int>mp;
mp['A']++;
mp['E']++;
mp['I']++;
mp['O']++;
mp['U']++;
mp['Y']++;
mp['a']++;
mp['e']++;
mp['i']++;
mp['o']++;
mp['u']++;
mp['y']++;
string str;
getline(cin,str);
char ch;
for(int i=str.length()-;i>=;i--){
if((str[i]>='A'&&str[i]<='Z')||(str[i]>='a'&&str[i]<='z')){
ch=str[i];
break;
}
}
if(mp[ch]) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
B
模拟进制转换和运算
#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define maxn 1000005
typedef long long ll;
typedef unsigned long long ull; int getmax(int a){
int Max=;
while(a){
Max=max(a%,Max);
a/=;
}
return Max;
} int Change(int n,int base){
int p=;
int ans=;
while(n){
ans=ans+(n%)*p;
n/=;
p*=base;
}
return ans;
} int main(){
#ifndef ONLINE_JUDGE
// freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
int a,b;
cin>>a>>b;
int base=max(getmax(a),getmax(b))+;
int sum=Change(a,base)+Change(b,base);
int ans=;
while(sum){
sum/=base;
ans++;
}
cout<<ans<<endl;
}
C
找规律
#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define maxn 1000005
typedef long long ll;
typedef unsigned long long ull; int main(){
#ifndef ONLINE_JUDGE
// freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
int n;
cin>>n;
cout<<n<<" ";
for(int i=;i<n;i++) cout<<i<<" ";
}
D
枚举第一个是0还是1,然后不断向后遍历判断,取最小值
#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define maxn 1000005
typedef long long ll;
typedef unsigned long long ull; int n;
string str; int func(int ch){
int ans=;
if(str[]!=ch+'') ans++;
ch^=;
for(int i=;i<str.length();i++){
if(str[i]!=ch+''){
ans++;
}
ch^=;
}
return ans;
} int main(){
#ifndef ONLINE_JUDGE
// freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
cin>>n;
cin>>str;
int ans=0x3f3f3f3f;
ans=min(ans,func());
ans=min(ans,func());
cout<<ans<<endl;
}
E
区间DP
参考博客:https://blog.csdn.net/zhjchengfeng5/article/details/8201105
#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 1000005
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef unsigned long long ull; string s[];
int n;
string str[];
int dp[][];
bool book[][][][];
int len[]; void Init(int id){
len[id]=s[id].length();
rep(i,,len[id]){
book[id][i][i][s[id][i]-'a']=;
}
rep(L,,len[id]+){
int st=;
rep(en,st+L-,len[id]){
rep(mid,st,en){
rep(i,,n){
if(book[id][st][mid][str[i][]-'a']&&book[id][mid+][en][str[i][]-'a']){
book[id][st][en][str[i][]-'a']=;
}
}
}
st++;
}
}
} int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
cin>>s[]>>s[];
cin>>n;
for(int i=;i<n;i++){
cin>>str[i];
}
Init(),Init();
memset(dp,0x3f,sizeof(dp));
dp[][]=;
rep(mid0,,len[]){
rep(mid1,,len[]){
rep(st0,mid0,len[]){
rep(st1,mid1,len[]){
rep(ch,,){
if(book[][mid0][st0][ch]&&book[][mid1][st1][ch]){
dp[st0+][st1+]=min(dp[st0+][st1+],dp[mid0][mid1]+);
}
}
}
}
}
}
int ans=dp[len[]][len[]];
if(ans==0x3f3f3f3f) cout<<-<<endl;
else cout<<ans<<endl;
}
Codeforces Beta Round #46 (Div. 2)的更多相关文章
- Codeforces Beta Round #80 (Div. 2 Only)【ABCD】
Codeforces Beta Round #80 (Div. 2 Only) A Blackjack1 题意 一共52张扑克,A代表1或者11,2-10表示自己的数字,其他都表示10 现在你已经有一 ...
- Codeforces Beta Round #83 (Div. 1 Only)题解【ABCD】
Codeforces Beta Round #83 (Div. 1 Only) A. Dorm Water Supply 题意 给你一个n点m边的图,保证每个点的入度和出度最多为1 如果这个点入度为0 ...
- Codeforces Beta Round #79 (Div. 2 Only)
Codeforces Beta Round #79 (Div. 2 Only) http://codeforces.com/contest/102 A #include<bits/stdc++. ...
- Codeforces Beta Round #77 (Div. 2 Only)
Codeforces Beta Round #77 (Div. 2 Only) http://codeforces.com/contest/96 A #include<bits/stdc++.h ...
- Codeforces Beta Round #76 (Div. 2 Only)
Codeforces Beta Round #76 (Div. 2 Only) http://codeforces.com/contest/94 A #include<bits/stdc++.h ...
- Codeforces Beta Round #75 (Div. 2 Only)
Codeforces Beta Round #75 (Div. 2 Only) http://codeforces.com/contest/92 A #include<iostream> ...
- Codeforces Beta Round #74 (Div. 2 Only)
Codeforces Beta Round #74 (Div. 2 Only) http://codeforces.com/contest/90 A #include<iostream> ...
- Codeforces Beta Round #73 (Div. 2 Only)
Codeforces Beta Round #73 (Div. 2 Only) http://codeforces.com/contest/88 A 模拟 #include<bits/stdc+ ...
- Codeforces Beta Round #72 (Div. 2 Only)
Codeforces Beta Round #72 (Div. 2 Only) http://codeforces.com/contest/84 A #include<bits/stdc++.h ...
随机推荐
- react-native启动页面设置,react-native-splash-screen
用于解决iOS和Android启动白屏问题及简单的启动页面展示 下载 react-native-splash-screen yarn add react-native-splash-screen re ...
- 使用Prometheus+Grafana监控MySQL实践
一.介绍Prometheus Prometheus(普罗米修斯)是一套开源的监控&报警&时间序列数据库的组合,起始是由SoundCloud公司开发的.随着发展,越来越多公司和组织接受采 ...
- CSS属性组-动画、转换、渐变
一.动画 animation动画属性是一个简写属性,用于设置六个动画属性 aninmation-name动画名称,被调用 animation-duration完成动画需要的持续时间 animation ...
- IIS Express
当前程序员只能通过下面两种web服务器之一来开发和测试ASP.NET网站程序: 1. Visual Studio自带的ASP.NET开发服务器(webdev.exe). 2. Windows自带的II ...
- sql help cs
using System;using System.Collections.Generic;using System.Linq;using System.Text;using System.Confi ...
- SQLServer 的数据分页:
假设现在有这样的一张表:CREATE TABLE test( id int primary key not null identity, names varchar(20))然后向里面插入大约1000 ...
- python函数基础:调用内置函数&定义函数
调用内置函数 有很多内置函数,在使用中需要积累.这里只举两个例子: 分别调用abs和数据类型转换,注意当入参类型错误时候会报错 ''' print('abs(-100)') abs(-100) pri ...
- windows 10 专业版 激活
参考文章:https://jingyan.baidu.com/article/c14654134b99de0bfcfc4c8c.html http://www.windowszj.com/news/2 ...
- asp.net中的reportview报错跟预编有关系
当报表控件出现: 报表定义无效.详细信息:根级别上的数据无效.行1,位置1. 先检查一下,你的aspx文件是不是变成了这样一句话 这是预编译工具生成的标记文件,不应被删除! 如果这样的话,报表控件是不 ...
- VC++ MFC如何生成一个可串行化的类
一.MFC允许对象在程序运行的整个过程中持久化的串行化机制(1)串行化是指向持久化存储媒介(如一个磁盘文件)读或写对象的过程.(2)串行化用于在程序运行过程时或之后修复结构化数据(如C++类或结构)的 ...