Codeforces Beta Round #73 (Div. 2 Only)

http://codeforces.com/contest/88

A

模拟

 #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 1000006
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull; string s[]={"C","C#","D","D#","E","F","F#","G","G#","A","B","H"},a,b,c;
int x[]; int main(){
#ifndef ONLINE_JUDGE
// freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
cin >> a >> b >> c;
for(int i=;i<;i++){
if(a==s[i] || b==s[i] || c==s[i])x[i]=;
}
for(int i=;i<;i++){
if(x[i]){
if(x[(i+)%] && x[(i+)%]){
cout << "major" << endl;
return ;
}
if(x[(i+)%] && x[(i+)%]){
cout << "minor" << endl;
return ;
}
}
}
cout << "strange" << endl;
}

B

模拟

 #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 1000006
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull; string str;
map<char,int>mp;
string s[];
map<char,int>book;
vector<pii>ve; double dis(int a,int b,int c,int d){
return sqrt(sqr(a-c)+sqr(b-d));
} int main(){
#ifndef ONLINE_JUDGE
// freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
int n,m,len;
double x;
cin>>n>>m>>x;
for(int i=;i<n;i++){
cin>>s[i];
for(int j=;j<m;j++){
mp[s[i][j]]=;
if(s[i][j]=='S') ve.pb(make_pair(i,j));
}
}
cin>>len>>str;
int ans=;
pii tmp;
double dist;
for(int k=;k<ve.size();k++){
for(int i=;i<n;i++){
for(int j=;j<m;j++){
dist=dis(ve[k].first,ve[k].second,i,j);
if(dist<=x) {
book[s[i][j]]=;
}
}
}
}
int i;
for(i=;i<len;i++){
if((str[i]>='A'&&str[i]<='Z'&&!mp[str[i]+])||(str[i]>='A'&&str[i]<='Z'&&!mp['S'])||(str[i]>='a'&&str[i]<='z'&&!mp[str[i]])) {
ans=;
break;
}
else if(str[i]>='A'&&str[i]<='Z'&&mp['S']){
if(!book[str[i]+]) ans++;
}
}
if(i==len&&ans==) cout<<<<endl;
else if(!ans) cout<<-<<endl;
else cout<<ans<<endl;
}

C

gcd

 #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 1000006
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull; int main(){
#ifndef ONLINE_JUDGE
// freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
int a,b;
cin >> a>>b;
cout <<((abs(a/__gcd(a,b)-b/__gcd(a,b))==)?"Equal":(a<b?"Dasha":"Masha"))<<endl;
}

D

模拟

 #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 1000006
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<double,double>pdd;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull; map<string,int>mp;
int n,cou;
string a,b,c;
void solve()
{
cin>>b;
int ans=;
string d="";
for(int i=; i<b.size(); i++)
{
if(b[i]=='&') ans--;
else if(b[i]=='*') ans++;
else d+=b[i];
}
cou=mp[d];
if(cou>) cou+=ans;
else cou=;
b.clear();
} int main(){
#ifndef ONLINE_JUDGE
// freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
mp["void"]=;
cin>>n;
while(n--)
{
cin>>a;
if(a=="typedef")
{
solve();
cin>>c;
mp[c]=cou;
c.clear();
}
else if(a=="typeof")
{
solve();
if(--cou<) cout<<"errtype"<<endl;
else
{
cout<<"void";
for(int i=; i<cou; i++) cout<<"*";
cout<<endl;
}
}
a.clear();
}
}

E

sg函数(照着AC代码打的,没有完全理解)

 #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 maxn 100005
typedef long long ll;
typedef unsigned long long ull;
const ull MOD=;
/*#ifndef ONLINE_JUDGE
freopen("1.txt","r",stdin);
#endif */ int ans[maxn],sg[maxn],mex[maxn]; void getsg(int n){ for(int i=;i*(i+)/<=n;i++){
if((*n)%i==){
int t=*n/i-i+;
if((t&)||t<) continue;
t/=;
mex[sg[t-+i]^sg[t-]]=n;
if((sg[t-+i]^sg[t-])==)
if(ans[n]==-)
ans[n]=i;
}
}
sg[n]=-;
for(int i=;;i++){
if(mex[i]!=n){
sg[n]=i;
break;
}
}
sg[n]^=sg[n-];
return ;
} int main(){
#ifndef ONLINE_JUDGE
// freopen("1.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
int n;
cin>>n;
memset(ans,-,sizeof(ans));
for(int i=;i<=n;i++){
getsg(i);
}
cout<<ans[n]<<endl;
}

Codeforces Beta Round #73 (Div. 2 Only)的更多相关文章

  1. 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 ...

  2. 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 ...

  3. Codeforces Beta Round #79 (Div. 2 Only)

    Codeforces Beta Round #79 (Div. 2 Only) http://codeforces.com/contest/102 A #include<bits/stdc++. ...

  4. Codeforces Beta Round #77 (Div. 2 Only)

    Codeforces Beta Round #77 (Div. 2 Only) http://codeforces.com/contest/96 A #include<bits/stdc++.h ...

  5. Codeforces Beta Round #76 (Div. 2 Only)

    Codeforces Beta Round #76 (Div. 2 Only) http://codeforces.com/contest/94 A #include<bits/stdc++.h ...

  6. Codeforces Beta Round #75 (Div. 2 Only)

    Codeforces Beta Round #75 (Div. 2 Only) http://codeforces.com/contest/92 A #include<iostream> ...

  7. Codeforces Beta Round #74 (Div. 2 Only)

    Codeforces Beta Round #74 (Div. 2 Only) http://codeforces.com/contest/90 A #include<iostream> ...

  8. Codeforces Beta Round #72 (Div. 2 Only)

    Codeforces Beta Round #72 (Div. 2 Only) http://codeforces.com/contest/84 A #include<bits/stdc++.h ...

  9. Codeforces Beta Round #70 (Div. 2)

    Codeforces Beta Round #70 (Div. 2) http://codeforces.com/contest/78 A #include<bits/stdc++.h> ...

随机推荐

  1. Yum安装MySQL以及相关目录路径和修改目录

    有些时候,为了方便,有些同学喜欢通过yum的方式安装MySQL,没有设置统一的文件目录以及软件目录,那么就会为后续的维护工作带来很大的麻烦! 下面就简单介绍一下yum安装MySQL的步骤以及这类安装下 ...

  2. nginx压缩,缓存

    https://www.darrenfang.com/2015/01/setting-up-http-cache-and-gzip-with-nginx/ https://www.linuxdashe ...

  3. html 基础之a标签的属性target解析

    学习前端,有很多标签其实有很多不同的功能,但是用到的不多,所以就没有发现:当发现的时候,觉得很不可思议,有耳目一新的感觉.例如a 标签,之前只是知道,使用a标签,可以打开一个链接,然后访问一个新的页面 ...

  4. 尚硅谷springboot学习22-Thymeleaf入门

    Thymeleaf是一种模板引擎,类似于JSP.Velocity.Freemarker

  5. linux下svn不能连接上windows服务器:SSL handshake failed: SSL error

    在linux服务器下载https链接的svn源码时出现:SSL handshake failed: SSL error: Key usage violation in certificate has ...

  6. U盘无法访问

    U盘无法访问 方法/步骤   首先,Win+R,打开“运行”窗口.   在打开的运行窗口中,输入cmd回车     这时会打开这样的一个窗口   这时输入chkdsk g: /f 需要说明的是,g这个 ...

  7. Java泛型类型擦除以及类型擦除带来的问题

    目录 1.Java泛型的实现方法:类型擦除 1-2.通过两个例子证明Java类型的类型擦除 2.类型擦除后保留的原始类型 3.类型擦除引起的问题及解决方法 3-1.先检查,再编译以及编译的对象和引用传 ...

  8. [C语言]变量VS常量

    -------------------------------------------------------------------------------------------- 1. 固定不变 ...

  9. Eclipse编译Android项目时出现的问题:Android requires compiler compliance level 5.0 or 6.0. Found '1.8' instead.

    Consle: Android requires compiler compliance level 5.0 or 6.0. Found '1.8' instead. Please use Andro ...

  10. 获取jdk支持的编码类型

    //获取jdk支持的编码类型 Map<String,Charset> maps = Charset.availableCharsets(); for(Map.Entry<String ...