Codeforces Beta Round #61 (Div. 2)
Codeforces Beta Round #61 (Div. 2)
http://codeforces.com/contest/66
A
输入用long double
#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 rep(k,i,j) for(int k=i;k<j;k++)
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);
long double n;
cin>>n;
if(n>=-&&n<=) cout<<"byte"<<endl;
else if(n>=-&&n<=) cout<<"short"<<endl;
else if(n>=-&&n<=) cout<<"int"<<endl;
else if(n<) cout<<"long"<<endl;
else{
cout<<"BigInteger"<<endl;
}
}
B
暴力枚举每一个数即可
#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 rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef unsigned long long ull; int a[]; int main(){
#ifndef ONLINE_JUDGE
// freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
int n;
cin>>n;
for(int i=;i<=n;i++){
cin>>a[i];
}
int ans=;
for(int i=;i<=n;i++){
int j=i-;
int co=;
while(j>=){
if(a[j+]>=a[j]){
j--;
co++;
}
else{
break;
}
}
j=i+;
while(j<=n){
if(a[j-]>=a[j]){
j++;
co++;
}
else{
break;
}
}
if(co>ans) ans=co;
}
cout<<ans<<endl;
}
C
模拟题
#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 rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef unsigned long long ull; string s,t;
int res1,res2;
map<string,int> M,N; int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
while(cin>>t)
{
s=t;
int F=;
while()
{
int x=s.find_last_of('\\');
if (x==) break;
s=s.substr(,x);
int f=N[s];
M[s]+=F;
N[s]++;
res1=max(res1,M[s]);
res2=max(res2,N[s]);
if (!f) F++;
}
}
cout<<res1<<' '<<res2<<endl;
}
D
找出3个数,使他们两两的公约数互不为1,他们三个的公约数为1,剩下的数就输出他们的倍数即可
#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 rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef unsigned long long ull; int a[]={,,}; int main(){
#ifndef ONLINE_JUDGE
// freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
int n;
cin>>n;
if(n==){
cout<<-<<endl;
}
else{
for(int i=;i<n;i++){
if(i<){
cout<<a[i]<<endl;
}
else{
cout<<*i<<endl;
}
}
}
}
E
找出a[i]-b[i]前缀和的最小值,然后依次减去,如果发现minn大于等于0的情况,说明走的通,逆向同理
#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 rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef unsigned long long ull; int n;
int a[];
int b[];
set<int>se;
set<int>::iterator it; void func(int x){
int minn=a[]-b[],pre=minn;
for(int i=;i<n;i++){
pre+=a[i]-b[i];
minn=min(minn,pre);
}
for(int i=;i<n;i++){
if(minn>=){
if(x){
se.insert(i+);
}
else{
se.insert(n-i);
}
}
minn-=a[i]-b[i];
}
} int main(){
#ifndef ONLINE_JUDGE
// freopen("input.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
cin>>n;
rep(i,,n) cin>>a[i];
rep(i,,n) cin>>b[i];
func();
reverse(a,a+n);
reverse(b,b+n-);
func();
cout<<se.size()<<endl;
for(it=se.begin();it!=se.end();it++){
cout<<*it<<" ";
}
}
Codeforces Beta Round #61 (Div. 2)的更多相关文章
- Codeforces Beta Round #61 (Div. 2) D. Petya and His Friends 想法
D. Petya and His Friends time limit per test 2 seconds memory limit per test 256 megabytes input sta ...
- Codeforces Beta Round #57 (Div. 2)
Codeforces Beta Round #57 (Div. 2) http://codeforces.com/contest/61 A #include<bits/stdc++.h> ...
- 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> ...
随机推荐
- ASP.net 完整登录流程
登录流程 using System; using System.Collections.Generic; using System.Linq; using System.Web; using Syst ...
- 用递归方法计算斐波那契数列第n项的和
参考: https://blog.csdn.net/xuzhangze/article/details/78568702 波那契数列数列从第3项开始,每一项都等于前两项之和.即 第n项的值为 (n- ...
- react-native-echarts 安卓版打包后,部分手机图表不显示问题
1. 找到 node_modules\native-echarts\src\components\Echarts\tpl.html 文件 ,把它复制到 (android\app\src\main\a ...
- 尚硅谷springboot学习5-主入口类说明
package com.atguigu; import org.springframework.boot.SpringApplication; import org.springframework.b ...
- ThinkPHP5分页样式设置
手册上讲分页类的使用时对样式讲的不够详细,这里我结合个人的摸索给大家一些参考意见. config里的分页配置我使用的是系统默认的bootstrap,查看thinkphp\library\think\p ...
- 数据结构:Queue
Queue设计与实现 Queue基本概念 队列是一种特殊的线性表 队列仅在线性表的两端进行操作 队头(Front):取出数据元素的一端 队尾(Rear):插入数据元素的一端 队列不允许在中间部位进行操 ...
- RunAsAdmin
program AdminCMD; {$APPTYPE CONSOLE} uses Windows, ShellApi, SysUtils; function RunAsAdmin(const ...
- python中的全局变量和局部变量(转)
python中,对于变量作用域的规定有些不一样. 在诸如C/C++.java等编程语言中,默认在函数的内部是能够直接訪问在函数外定义的全局变量的,可是这一点在python中就会有问题.以下是一个样例. ...
- How to fix the bug “Expected "required", "optional", or "repeated".”?
参考:https://github.com/tensorflow/models/issues/1834 You need to download protoc version 3.3 (already ...
- 对于“2017面向对象程序设计(Java)第五周工作总结”存在问题的反馈及本周教学计划
一:问题反馈 “上周我们学习的新内容主要是第五章,并对第四章内容做了巩固.从学生上交的实验报告完成情况以及学习Java心得博客中的反馈可以看出,学生对构造器.重载.超类.多态.抽象类这几个概念理解的不 ...