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

  1. 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 ...

  2. Codeforces Beta Round #57 (Div. 2)

    Codeforces Beta Round #57 (Div. 2) http://codeforces.com/contest/61 A #include<bits/stdc++.h> ...

  3. 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 ...

  4. 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 ...

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

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

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

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

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

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

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

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

  9. 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

  1. 【Source Insight 】之marco学习笔记2

    现在我们看先看一个 官方地址https://www.sourceinsight.com/download/macro-files/中的 autoexp.em Automatically expands ...

  2. TFDStoredProc执行sql server的部分存储过程报错,有的是好的。

    TFDStoredProc执行sql server的部分存储过程报错,有的是好的. Invalid character value for cast specification 暂时无解.用fdque ...

  3. js ajax 数据获取

    在js中应用ajax 获取数据的方法,也写一个出来供复习所用 1.建议一个user.json 文件如下,保存名字为 user.json { "name": "huanyi ...

  4. vue:在router里面给页面加title

    vue中给组件页面加页面标题:{ path: '/', name: 'index', component: disconnect, meta: { title: '首页' } }, { path: ' ...

  5. 如何安全的在不同工程间安全地迁移asset数据?三种方法

    答:1.将Assets和Library一起迁移2.导出包package3.用unity自带的assets Server功能

  6. html页面跳转

    button <button onclick="window.location.href='edit.jsp'">完善个人信息</button> 单击提交表 ...

  7. [福大2018高级软工教学]团队Beta阶段成绩汇总

    一.作业地址: https://edu.cnblogs.com/campus/fzu/AdvancedSoftwareEngineerning2018/homework/2465 二.Beta阶段作业 ...

  8. [CI]CodeIgniter应用配置明细

    ---------------------------------------------------------------------------------------------------- ...

  9. 使用django实现自定义用户认证

    参考资料:https://docs.djangoproject.com/en/1.10/topics/auth/customizing/    直接拉到最后看栗子啦 django自定义用户认证(使用自 ...

  10. ArcGIS案例学习笔记2_1_山顶点提取最大值提取

    ArcGIS案例学习笔记2_1_山顶点提取最大值提取 计划时间:第二天上午 目的:最大值提取 教程:Pdf page=343 数据:chap8/ex5/dem.tif 背景知识:等高线种类 基本等高线 ...