Educational Codeforces Round 65 (Rated for Div. 2)
A：签到。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 110
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int T,n;
char s[N];
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
T=read();
while (T--)
{
n=read();
scanf("%s",s+1);
int first=n;
for (int i=1;i<=n;i++) if (s[i]=='8') {first=i;break;}
if (n-first+1>=11) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
//NOTICE LONG LONG!!!!!
}
B：太难了吧。注意到任意两数乘积不同。于是考虑问出12乘积、34乘积，由此已经可知56乘积。然后需要确定每一对的顺序，可以询问13得到13的值，再询问15得到5的值。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 110
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int w[7]={0,4,8,15,16,23,42},a[7];
bool isac(int x)
{
for (int i=1;i<=6;i++) if (w[i]==x) return 1;
return 0;
}
signed main()
{
int x,y,z;
cout<<"?"<<' '<<1<<' '<<2<<endl;x=read();
cout<<"?"<<' '<<3<<' '<<4<<endl;y=read();
cout<<"?"<<' '<<1<<' '<<3<<endl;z=read();
cout<<"?"<<' '<<1<<' '<<5<<endl;int t=read();
for (int i=1;i<=6;i++)
{
if (x%w[i]==0&&z%w[i]==0&&t%w[i]==0&&isac(x/w[i])&&isac(z/w[i])&&isac(t/w[i]))
{
a[1]=w[i];
a[2]=x/w[i];
a[3]=z/w[i];
a[5]=t/w[i];
}
}
a[4]=y/a[3];
a[6]=1;for (int i=1;i<=6;i++) a[6]*=w[i];
for (int i=1;i<=5;i++) a[6]/=a[i];
cout<<"!"<<' ';for (int i=1;i<=6;i++) cout<<a[i]<<' ';
return 0;
//NOTICE LONG LONG!!!!!
}
C：并查集。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 500010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,a[N],fa[N],size[N];
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
signed main()
{
n=read(),m=read();
for (int i=1;i<=n;i++) fa[i]=i;
for (int i=1;i<=m;i++)
{
int t=read();
for (int j=1;j<=t;j++)
{
a[j]=read();
}
for (int j=1;j<t;j++)
fa[find(a[j])]=find(a[j+1]);
}
for (int i=1;i<=n;i++) size[find(i)]++;
for (int i=1;i<=n;i++) printf("%d ",size[find(i)]);
return 0;
//NOTICE LONG LONG!!!!!
}
D：贪心。左括号分给当前前缀和小的，右括号分给当前前缀和大的。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,x,y,a[N];
char s[N];
bool flag[N];
signed main()
{
n=read();
scanf("%s",s+1);
for (int i=1;i<=n;i++) if (s[i]=='(') a[i]=1;else a[i]=-1;
for (int i=1;i<=n;i++)
if (a[i]==1)
{
if (x<y) x++,flag[i]=0;
else y++,flag[i]=1;
}
else
{
if (x>y) x--,flag[i]=0;
else y--,flag[i]=1;
}
for (int i=1;i<=n;i++) printf("%d",flag[i]);
return 0;
//NOTICE LONG LONG!!!!!
}
E：对于确定的右端点r，显然合法的l是一段前缀。考虑先求出令保留[1,l]合法的最大l和令保留[r,m]合法的最小r，然后在范围内twopointers，记一下小于l的值的最晚出现位置和大于r的值的最早出现位置即可。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 1000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,a[N],mx[N];
vector<int> pos[N];
set<int> q;
ll ans;
signed main()
{
n=read(),m=read();
for (int i=1;i<=n;i++) a[i]=read();
for (int i=1;i<=n;i++) pos[a[i]].push_back(i);
int p=n+1;int l=m,r=1;
for (int i=m;i>=1;i--)
{
bool flag=0;
int x=n+1;
for (auto j:pos[i])
{
if (j>p) {flag=1;break;}
x=min(x,j);
}
p=min(p,x);
if (flag) {r=i;break;}
}
p=0;
for (int i=1;i<=m;i++)
{
bool flag=0;
int x=0;
for (auto j:pos[i])
{
if (j<p) {flag=1;break;}
x=max(x,j);
}
p=max(p,x);
if (flag) {l=i;break;}
}
int x=1;int y=0;
for (int i=r;i<=m;i++)
for (auto j:pos[i]) q.insert(j);
for (int i=1;i<=m;i++)
{
for (auto j:pos[i])
mx[i]=max(mx[i],j);
}
q.insert(n+1);
for (int i=r;i<=m;i++)
{
for (auto j:pos[i]) q.erase(j);
while (x<i&&x<l&&max(y,mx[x])<*q.begin()) y=max(y,mx[x++]);
ans+=x;
}
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
F：考虑每个数的贡献。区间内每有一个数比它小它的贡献次数就+1，初始贡献次数为1。于是考虑所有其他数对该数贡献次数的贡献。对于处于ax和ay的值，显然包含其的区间个数为x*(n-y+1)。于是要求出所有在其前方比它小的数的位置和与所有在其后方比它小的数的位置和。按权值从小到大考虑每个数，树状数组维护即可。wa了两发49简直自闭。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define int long long
#define N 500010
#define P 1000000007
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,a[N],b[N],p[N],tree2[N];
ll ans,s,tree[N];
void ins(int x){int k=x;while (k<=n) tree[k]+=x,k+=k&-k;}
int query(int k){ll s=0;while (k) s+=tree[k],k-=k&-k;return s%P;}
void ins2(int x){while (x<=n) tree2[x]++,x+=x&-x;}
int query2(int k){int s=0;while (k) s+=tree2[k],k-=k&-k;return s;}
signed main()
{
n=read();
for (int i=1;i<=n;i++) b[i]=a[i]=read();
sort(b+1,b+n+1);
for (int i=1;i<=n;i++) a[i]=lower_bound(b+1,b+n+1,a[i])-b;
for (int i=1;i<=n;i++) p[a[i]]=i;
for (int i=1;i<=n;i++)
{
int x=b[i];
ans=(ans+1ll*x*(1ll*p[i]*(n-p[i]+1)%P+1ll*query(p[i])*(n-p[i]+1)%P+1ll*(1ll*(i-1-query2(p[i]))*(n+1)%P+P-(s+P-query(p[i]))%P)%P*p[i]%P)%P)%P;
ins(p[i]);ins2(p[i]);s+=p[i];
}
cout<<ans%P;
return 0;
//NOTICE LONG LONG!!!!!
}
G：咕了咕了。
小小小号。result：rank 18 rating +239
Educational Codeforces Round 65 (Rated for Div. 2)的更多相关文章
- Educational Codeforces Round 65 (Rated for Div. 2)题解
Educational Codeforces Round 65 (Rated for Div. 2)题解 题目链接 A. Telephone Number 水题,代码如下: Code #include ...
- Educational Codeforces Round 65 (Rated for Div. 2) D. Bicolored RBS
链接:https://codeforces.com/contest/1167/problem/D 题意: A string is called bracket sequence if it does ...
- Educational Codeforces Round 65 (Rated for Div. 2) C. News Distribution
链接:https://codeforces.com/contest/1167/problem/C 题意: In some social network, there are nn users comm ...
- Educational Codeforces Round 65 (Rated for Div. 2) B. Lost Numbers
链接:https://codeforces.com/contest/1167/problem/B 题意: This is an interactive problem. Remember to flu ...
- Educational Codeforces Round 65 (Rated for Div. 2) A. Telephone Number
链接:https://codeforces.com/contest/1167/problem/A 题意: A telephone number is a sequence of exactly 11 ...
- Educational Codeforces Round 65 (Rated for Div. 2)B. Lost Numbers（交互）
This is an interactive problem. Remember to flush your output while communicating with the testing p ...
- [ Educational Codeforces Round 65 (Rated for Div. 2)][二分]
https://codeforc.es/contest/1167/problem/E E. Range Deleting time limit per test 2 seconds memory li ...
- Educational Codeforces Round 65 (Rated for Div. 2) E. Range Deleting（思维+coding）
传送门 参考资料: [1]:https://blog.csdn.net/weixin_43262291/article/details/90271693 题意: 给你一个包含 n 个数的序列 a,并且 ...
- Educational Codeforces Round 65 (Rated for Div. 2)(ACD)B是交互题，不怎么会
A. Telephone Number A telephone number is a sequence of exactly 11 digits, where the first digit is ...
随机推荐
- 《Glibc内存管理》笔记DAY5
目录 分箱式内存管理 Unsorted bin Fast bins 核心结构体分析 malloc_state 内容来源 分箱式内存管理 Unsorted bin Unsorted bin 可以看作 ...
- 解决问题：OSError: mysql_config not found
通过pip install mysqlclient时报出了OSError: mysql_config not found错误,如下 Traceback (most recent call last): ...
- es6对象复制合并 Object.assign
对象的复制 var obj= { a: 1 }; var copy = Object.assign({}, obj); console.log(copy); //{ a: 1 } 对象的合并和封装 v ...
- Android：Mstar平台 HDMI OUT 静音流程
一.framework层 1. APP调用 AudioManager 的 adjustStreamVolume() 接口实现在: frameworks\base\services\core\java\ ...
- MySQL数据库备份之xtrabackup工具使用
一.Xtrabackup备份介绍及原理 二.Xtrabackup的安装 1.在centos7上基于yum源安装percona-xtrabackup软件 [root@node7 ~]# yum -y i ...
- myeclipse打开jsp后卡死的问题详解
myeclipse打开jsp后卡死的问题详解 1,打开 Window -> Preferences -> General -> Editors -> File Associ ...
- 取用户中文名 FDM_CUST_USER_NAME_READ_SINGLE
DATA:lv_first TYPE ad_namefir, lv_last TYPE ad_namelas, lv_full TYPE ad_namtext. CALL ...
- Pyhthon3之使用__slots__
正常情况下,我们定义了一个class,创建了一个class实例后,我们可以给该实例绑定任何属性和方法,这就是动态语言的灵活性.先定义class: >>> class Student( ...
- 第二十一章 授予身份及切换身份——《跟我学Shiro》
目录贴:跟我学Shiro目录贴 在一些场景中,比如某个领导因为一些原因不能进行登录网站进行一些操作,他想把他网站上的工作委托给他的秘书,但是他不想把帐号/密码告诉他秘书,只是想把工作委托给他:此时和我 ...
- python：序列化与反序列化（json、pickle、shelve）
本节内容 前言 json模块 pickle模块 shelve模块 总结 一.前言 1. 现实需求 每种编程语言都有各自的数据类型,其中面向对象的编程语言还允许开发者自定义数据类型(如:自定义类),Py ...