第一题水题,8分钟1a

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f3f; int main()
{
ios::sync_with_stdio(false);
cin.tie();
int n;
cin>>n;
int last;
bool maybe=,rate=;
for(int i=;i<n;i++)
{
int a,b;
cin>>a>>b;
if(i==)last=a;
else
{
if(last<a)maybe=;
last=a;
}
if(a!=b)rate=;
}
if(rate)cout<<"rated"<<endl;
else if(!rate&&maybe)cout<<"unrated"<<endl;
else if(!rate&&!maybe)cout<<"maybe"<<endl;
return ;
}

A

第二题暴力模拟一下,每次加(减)50,先减再加

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f3f; int main()
{
ios::sync_with_stdio(false);
cin.tie();
int p,x,y,ans=;
cin>>p>>x>>y;
for(int i=x;i>=y;i-=)
{
int j=(i/)%;
for(int k=;k<=;k++)
{
j=(j*+)%;
if(j+==p)
{
ans=i;
break;
}
}
if(ans)break;
}
if(ans)
{
cout<<<<endl;
return ;
}
for(int i=x;;i+=)
{
int j=(i/)%;
for(int k=;k<=;k++)
{
j=(j*+)%;
if(j+==p)
{
ans=i;
break;
}
}
if(ans)break;
}
int res=(ans-x+)/;
cout<<res<<endl;
return ;
}

B

第三题wa了6发,终于O(1)过了,数据1e9早该想到算法有问题

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f3f; int main()
{
ios::sync_with_stdio(false);
cin.tie();
ll x,y,p,q;
int t;
cin>>t;
while(t--){
cin>>x>>y>>p>>q;
if(p==q)
{
if(x==y)cout<<<<endl;
else cout<<-<<endl;
continue;
}
if(p==)
{
if(x==)cout<<<<endl;
else cout<<-<<endl;
continue;
}
ll k1=(y-x)/(q-p);
if((y-x)%(q-p)!=)k1++;
ll k2=x/p;
if(x%p!=)k2++;
ll k3=y/q;
if(y%q!=)k3++;
ll k=max(k1,max(k2,k3));
cout<<k*q-y<<endl;
}
return ;
}

C

Codeforces Round #412的更多相关文章

  1. Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3)(A.B.C,3道暴力题,C可二分求解)

    A. Is it rated? time limit per test:2 seconds memory limit per test:256 megabytes input:standard inp ...

  2. Codeforces Round#412 Div.2

    A. Is it rated? 题面 Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codefo ...

  3. Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3) A B C D 水 模拟 二分 贪心

    A. Is it rated? time limit per test 2 seconds memory limit per test 256 megabytes input standard inp ...

  4. Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3) D - Dynamic Problem Scoring

    地址:http://codeforces.com/contest/807/problem/D 题目: D. Dynamic Problem Scoring time limit per test 2 ...

  5. Codeforces Round #412 A Is it rated ?

    A. Is it rated? time limit per test  2 seconds memory limit per test  256 megabytes Is it rated? Her ...

  6. Codeforces Round #412 Div. 2 第一场翻水水

    大半夜呆在机房做题,我只感觉智商严重下降,今天我脑子可能不太正常 A. Is it rated? time limit per test 2 seconds memory limit per test ...

  7. Codeforces Round #412 Div. 2 补题 D. Dynamic Problem Scoring

    D. Dynamic Problem Scoring time limit per test 2 seconds memory limit per test 256 megabytes input s ...

  8. Codeforces Round #412 B. T-Shirt Hunt

    B. T-Shirt Hunt time limit per test 2 seconds memory limit per test 256 megabytes input standard inp ...

  9. Codeforces Round #412 C. Success Rate

    C. Success Rate time limit per test 2 seconds memory limit per test 256 megabytes input standard inp ...

  10. Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3) A Is it rated?

    地址:http://codeforces.com/contest/807/problem/C 题目: C. Success Rate time limit per test 2 seconds mem ...

随机推荐

  1. JS根据userAgent值来判断浏览器的类型及版本【转】

    转自:http://blog.csdn.net/sunlovefly2012/article/details/22384255 JavaScript是前端开发的主要语言,我们可以通过编写JavaScr ...

  2. cpu与寄存器,内核态与用户态及如何切换

    cpu:相当于计算机的大脑负责运算和发送命令: 寄存器:寄存器是cpu当中的一个有限存储部件,cpu从内存调用数据时,寄存器会将从内存调用的数据进行更新在寄存器中以一个字或变量进行存储. 寄存器总共分 ...

  3. php写守护进程(转载 http://blog.csdn.net/tengzhaorong/article/details/9764655)

    守护进程(Daemon)是运行在后台的一种特殊进程.它独立于控制终端并且周期性地执行某种任务或等待处理某些发生的事件.守护进程是一种很有用的进程.php也可以实现守护进程的功能. 1.基本概念 进程 ...

  4. 【android】开发笔记系列UI篇

    弹出View添加阴影效果 系统自带就有,在android studio上直接写入背景颜色 android:background="@android:drawable/dialog_holo_ ...

  5. Java 为什么要使用反射(通俗易懂的举例)

    Java反射最大的好处就是能在运行期间,获得某个类的结构.成员变量,用来实例化. 下列是具体使用场景:假如我们有两个程序员,一个程序员在写程序的时候,需要使用第二个程序员所写的类,但第二个程序员并没完 ...

  6. ReentrantLock的底层实现机制 AQS

    ReentrantLock的底层实现机制是AQS(Abstract Queued Synchronizer 抽象队列同步器).AQS没有锁之类的概念,它有个state变量,是个int类型,为了好理解, ...

  7. c++对txt文件的读取与写入

    转自:http://blog.csdn.net/lh3325251325/article/details/4761575 #include <iostream> #include < ...

  8. hive union all使用注意

    UNION用于联合多个select语句的结果集,合并为一个独立的结果集,结果集去重. UNION ALL也是用于联合多个select语句的结果集.但是不能消除重复行.现在hive只支持UNION AL ...

  9. windows安装git客户端

    1:线上git地址 https://github.com/ 2:tortoiseGit地址 http://tortoisegit.org 3:安装步骤 操作系统:Windows XP SP3 Git客 ...

  10. OpenStack之Keystone模块

    一.Keystone介绍 OpenStack Identity(Keystone)服务为运行OpenStack Compute上的OpenStack云提供了认证和管理用户.帐号和角色信息服务,并为Op ...