codeforces gym101243 A C D E F G H J
A
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#include<set>
#define ll __int64
#define mod 100000000
#define N 5e6+10
#define M 1e
using namespace std;
int n,k;
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
scanf("%d %d",&n,&k);
printf("%d\n",max((n*+k-)/k,)); return ;
}
C
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#include<set>
#define ll __int64
#define mod 100000000
#define N 5e6+10
#define M 1e
using namespace std;
int h,w;
int x[],y[];
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
int exm=;
scanf("%d %d",&h,&w);
if(h<=||w<=)
{
printf("0\n");
return ;
}
if(h%==){
for(int i=h-;i>=;i-=){
for(int j=;j<w;j++){
x[exm]=i;
y[exm]=j;
exm++;
}
if(i!=){
x[exm]=i-;
y[exm]=w-;
exm++;
}
}
}
else{
for(int i=h-;i>=;i-=){
for(int j=;j<w;j++){
x[exm]=i;
y[exm]=j;
exm++;
}
x[exm]=i-;
y[exm]=w-;
exm++;
}
for(int j=;j<w;j+=){
if(j!=w-){
x[exm]=;
y[exm]=j;
exm++;
x[exm]=;
y[exm]=j;
exm++;
x[exm]=;
y[exm]=j+;
exm++;
}
else{
x[exm]=;
y[exm]=j;
exm++;
x[exm]=;
y[exm]=j;
exm++;
}
}
}
printf("%d\n",exm);
for(int i=;i<exm;i++)
printf("%d %d\n",x[i],y[i]);
return ;
}
D
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
using namespace std;
#define PI acos(-1.0)
typedef long long ll;
typedef pair<int,int> P;
const int maxn=1e4+,maxm=1e5+,inf=0x3f3f3f3f,mod=1e9+;
const ll INF=1e13+;
struct edge
{
int from,to;
int cost;
};
edge es[maxm];
priority_queue<P,vector<P>,greater<P> >que;
char s[];
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
cin>>s;
int n=strlen(s);
int ans=;
for(int i=;i<n;i++)
{
if(s[i]=='E'&&(s[i-]=='N'||s[i-]=='S')) ans=(ans*)%mod;
else if(s[i]=='W'&&(s[i-]=='N'||s[i-]=='S')) ans=(ans*)%mod;
}
cout<<ans<<endl;
return ;
}
E
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#include<set>
#define ll __int64
#define mod 100000000
#define N 5e6+10
#define M 1e
using namespace std;
ll n,k,a[];
ll sum[];
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
scanf("%I64d %I64d",&n,&k);
sum[]=;
ll maxn=;
int pos;
for(ll i=; i<=n; i++)
{
scanf("%I64d",&a[i]);
sum[i]=sum[i-]+a[i];
if(maxn<a[i])
{
pos=i;
maxn=a[i];
}
}
ll l,r;
ll ans1,ans2;
l=n-+maxn;
r=sum[n];
ans1=max(0ll,k-pos+)/l;
ans2=max(0ll,k-sum[pos-])/r;
for(ll i=; i<=ans1; i++)
if(k>=(pos-+l*i)&&k<=(sum[pos-]+r*i))
{
printf("YES\n");
return ;
}
printf("KEK\n");
return ;
}
F
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
const int maxn=1e3+,maxm=1e6+,inf=0x3f3f3f3f,mod=1e9+;
const ll INF=1e13+;
struct edge
{
int from,to;
char ch;
};
edge es[maxm];
int pa[maxn];
int findset(int x)
{
return pa[x]==x?x:pa[x]=findset(pa[x]);
}
void unit(int x,int y)
{
int fx=findset(x),fy=findset(y);
if(fx<=fy) pa[fx]=fy;
else pa[fy]=fx;
}
int in[maxn],out[maxn];
int ans[maxn];
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
int n,k;
scanf("%d%d",&n,&k);
for(int i=; i<=n; i++) pa[i]=i;
memset(out,,sizeof(out));
memset(in,,sizeof(in));
for(int i=; i<=k; i++)
{
scanf("%d%c%d",&es[i].from,&es[i].ch,&es[i].to);
if(es[i].ch=='=') unit(es[i].from,es[i].to);
else
{
if(es[i].ch=='>') swap(es[i].from,es[i].to);
es[i].ch='<';
}
}
for(int i=; i<=k; i++)
{
if(es[i].ch=='=') continue;
out[findset(es[i].from)]++,in[findset(es[i].to)]++;
}
for(int i=; i<=n; i++)
if(in[i]&&out[i]) ans[i]=;
for(int i=; i<=k; i++)
{
if(es[i].ch=='=') continue;
if(ans[findset(es[i].from)]==) ans[findset(es[i].to)]=;
else if(ans[findset(es[i].to)]==) ans[findset(es[i].from)]=;
}
for(int i=; i<=n; i++)
{
if(ans[findset(i)]==) cout<<"B";
else if(ans[findset(i)]==) cout<<"R";
else if(ans[findset(i)]==) cout<<"W";
else cout<<"?";
}
cout<<endl;
return ;
}
G
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#include<set>
#define ll __int64
#define mod 100000000
#define N 5e6+10
#define M 1e
using namespace std;
int n;
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
scanf("%d",&n);
int que[];
int jishu=;
for(int i=;i*i<=n;i++)
{
if(n%i==)
{
int flag=;
que[jishu++]=i;
while(n%i==)
{
flag++;
n/=i;
if(flag>)
{
printf("NO\n");
return ;
}
}
}
}
if(n>)
que[jishu++]=n;
if(jishu==)
printf("YES\n");
else
printf("NO\n");
return ;
}
H
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#include<set>
#define ll __int64
#define mod 100000000
#define N 5e6+10
#define M 1e
using namespace std;
int n;
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
scanf("%d",&n);
if(n<)
{
ll ans=;
for(int i=;i<=n;i++)
ans*=;
printf("%I64d\n",ans);
}
else
{
ll ans=;
for(int i=;i<=;i++)
ans*=;
printf("%I64d",ans);
for(int i=;i<=n;i++)
printf("");
printf("\n");
}
return ;
}
J
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
const int maxn=1e3+,maxm=1e6+,inf=0x3f3f3f3f,mod=1e9+;
const ll INF=1e13+;
struct edge
{
int from,to;
char ch;
};
edge es[maxm];
char num[maxn][];
vector<int>G[maxn];
int cy[maxn],vis[maxn];
bool dfs(int u)
{
for(int i=; i<G[u].size(); i++)
{
int v=G[u][i];
if(vis[v]) continue;
vis[v]=true;
if(cy[v]==-||dfs(cy[v]))
{
cy[v]=u;
return true;
}
}
return false;
}
int solve(int n)
{
int ret=;
memset(cy,-,sizeof(cy));
for(int i=; i<=n; i++)
{
memset(vis,,sizeof(vis));
if(dfs(i)) ret++;
}
return ret;
}
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
int n,m;
scanf("%d%d",&n,&m);
for(int i=; i<=n; i++) scanf("%s",num[i]);
for(int i=; i<=m; i++) scanf("%s",num[i+n]);
for(int i=; i<=n; i++)
{
for(int j=; j<=m; j++)
{
if(num[i][]+num[i][]+num[i][]==num[j+n][]+num[j+n][]+num[j+n][])
G[i].push_back(j+n);
if(num[i][]+num[i][]+num[i][]==num[j+n][]+num[j+n][]+num[j+n][])
G[i].push_back(j+n);
}
}
cout<<solve(n)<<endl;
for(int i=n+; i<=n+m; i++)
{
if(cy[i]==-)continue;
if(num[cy[i]][]+num[cy[i]][]+num[cy[i]][]==num[i][]+num[i][]+num[i][])
cout<<"AT"<<" "<<num[cy[i]]<<" "<<num[i]<<endl;
else if(num[i][]+num[i][]+num[i][]==num[cy[i]][]+num[cy[i]][]+num[cy[i]][])
cout<<"TA"<<" "<<num[i]<<" "<<num[cy[i]]<<endl;
}
return ;
}
codeforces gym101243 A C D E F G H J的更多相关文章
- 2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest A E F G H I K M
// 深夜补水题,清早(雾)写水文 A. Automatic Door 题意 \(n(n\leq 1e9)\)个\(employee\)和\(m(m\leq 1e5)\)个\(client\)要进门, ...
- codeforces gym 100952 A B C D E F G H I J
gym 100952 A #include <iostream> #include<cstdio> #include<cmath> #include<cstr ...
- Codeforces 1214 F G H 补题记录
翻开以前打的 #583,水平不够场上只过了五题.最近来补一下题,来记录我sb的调试过程. 估计我这个水平现场也过不了,因为前面的题已经zz调了好久-- F:就是给你环上一些点,两两配对求距离最小值. ...
- Gym 101606 - A/B/C/D/E/F/G/H/I/J/K/L - (Undone)
链接:https://codeforces.com/gym/101606 A - Alien Sunset 暴力枚举小时即可. #include<bits/stdc++.h> using ...
- The 16th Zhejiang Provincial Collegiate Programming Contest Sponsored(E F G H I)
http://acm.zju.edu.cn/onlinejudge/showContestProblems.do?contestId=392 E:Sequence in the Pocket 思路:从 ...
- Codeforces Round #541 (Div. 2) (A~F)
目录 Codeforces 1131 A.Sea Battle B.Draw! C.Birthday D.Gourmet choice(拓扑排序) E.String Multiplication(思路 ...
- Educational Codeforces Round 39 (Rated for Div. 2) G
Educational Codeforces Round 39 (Rated for Div. 2) G 题意: 给一个序列\(a_i(1 <= a_i <= 10^{9}),2 < ...
- AtCoder Beginner Contest 220部分题(G,H)题解
刚开始的时候被E题卡住了,不过发现是个数学题后就开始使劲推式子,幸运的是推出来了,之后的F题更是树形DP换根的模板吧,就草草的过了,看了一眼G,随便口胡了一下,赶紧打代码,毕竟时间不多了,最后也没打完 ...
- //给定N个整数序列｛A1,A2,A3...An｝,求函数f（i,j）=（k=i~j）Ak的求和
//给定N个整数序列{A1,A2,A3...An},求函数f(i,j)=(k=i~j)Ak的求和 # include<stdio.h> void main() { ,sum1; ]={,- ...
随机推荐
- selenium 列表循环定位方法。
话不多说,直接上代码. 就是循环第一层,然后拼接,然后继续循环,继续屏接,任你多少层都不是问题. def c_select(self, values, text): """ ...
- linux下各文件夹的结构说明及用途介绍(转载)
详细介绍文档 转载文章路径 /bin:二进制可执行命令. /dev:设备特殊文件. /etc:系统管理和配置文件. /etc/rc.d:启动的配 置文件和脚本. /home:用户主目录的基点,比如用户 ...
- GitLab 搭建与使用
操作系统:Centos 7 环境:VM虚拟机 0x00:这里说下VM 虚拟机的配置 然后选择NAT模式 接下来配置网络 cd /etc/sysconfig/network-scripts/ 编辑:vi ...
- sparkSQL中RDD——DataFrame——DataSet的区别
spark中RDD.DataFrame.DataSet都是spark的数据集合抽象,RDD针对的是一个个对象,但是DF与DS中针对的是一个个Row RDD 优点: 编译时类型安全 编译时就能检查出类型 ...
- ES6中Class的继承关系
es5实现中,每个对象都有__proto__属性(也就是关系图中[[prototype]]属性),指向对应的构造函数的prototype.Class 作为构造函数的语法糖,同时有prototype属性 ...
- CentOS 6.7下创建桌面快捷方式
CentOS 6.7下创建桌面快捷方式如下: 1 在桌面右键,选择“创建启动器" 2 在弹出菜单中,填写名称(显示在桌面上的名字),命令(可执行程序的路径) 3 点击弹出菜单左边的图标,选择 ...
- 20135208JAVA第二次试验
北京电子科技学院(BESTI) 实 验 报 告 课程:Java程序设计 班级:1352 姓名:贺邦 学号:20135208 成绩: 指导教师:娄嘉鹏 ...
- 软工实践-Alpha 冲刺 （4/10）
队名:起床一起肝活队 组长博客:博客链接 作业博客:班级博客本次作业的链接 组员情况 组员1(队长):白晨曦 过去两天完成了哪些任务 描述: 很胖,刚学,照猫画虎做了登录与注册界面. 展示GitHub ...
- CentOS 7 安装 MySql 8
1-安装 CentOS 7 2-安装 NETCORE SDK SDK 安装文档:https://dotnet.microsoft.com/download/linux-package-m ...
- Codeforces Round #272 (Div. 2) E. Dreamoon and Strings dp
题目链接: http://www.codeforces.com/contest/476/problem/E E. Dreamoon and Strings time limit per test 1 ...