AtCoder Regular Contest 103 题解
C-/\/\/\
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<complex>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<ctime>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define Cpy(x,y) memcpy(x,y,sizeof(x))
#define Set(x,y) memset(x,y,sizeof(x))
#define FILE "a"
#define mp make_pair
#define pb push_back
#define RG register
#define il inline
using namespace std;
typedef unsigned long long ull;
typedef vector<int>VI;
typedef long long ll;
typedef double dd;
const int N=100010;
const int mod=19940417;
const int base=26;
const dd eps=1e-6;
const int inf=2147483647;
const ll INF=1ll<<60;
const ll P=100000;
il ll read(){
RG ll data=0,w=1;RG char ch=getchar();
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
return data*w;
}
il void file(){
srand(time(NULL)+rand());
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
}
int n,t[2][N],pre[2][N],suf[2][N],ans;
int main()
{
n=read();ans=n;
for(RG int i=1;i<=n;i++)t[i&1][read()]++;
for(RG int i=1;i<=100000;i++)
pre[0][i]=max(pre[0][i-1],t[0][i]),pre[1][i]=max(pre[1][i-1],t[1][i]);
for(RG int i=100000;i;i--)
suf[0][i]=max(suf[0][i+1],t[0][i]),suf[1][i]=max(suf[1][i+1],t[1][i]);
for(RG int i=1;i<=100000;i++)
ans=min(ans,n-(t[0][i]+max(pre[1][i-1],suf[1][i+1])));
printf("%d\n",ans);
return 0;
}
D-Robot Arms
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<complex>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<ctime>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define Cpy(x,y) memcpy(x,y,sizeof(x))
#define Set(x,y) memset(x,y,sizeof(x))
#define FILE "a"
#define mp make_pair
#define pb push_back
#define RG register
#define il inline
using namespace std;
typedef unsigned long long ull;
typedef vector<int>VI;
typedef long long ll;
typedef double dd;
const int N=100010;
const int mod=19940417;
const int base=26;
const dd eps=1e-6;
const int inf=2147483647;
const ll INF=1ll<<60;
const ll P=100000;
il ll read(){
RG ll data=0,w=1;RG char ch=getchar();
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
return data*w;
}
il void file(){
srand(time(NULL)+rand());
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
}
int n,m,x[N],y[N],p[N],q[N],k[N],s[N];
il void er(){puts("-1");exit(0);}
int main()
{
n=read();m=31;
for(RG int i=1;i<=n;i++){
x[i]=read();y[i]=read();
p[i]=x[i]+y[i];q[i]=x[i]-y[i];
x[i]=abs(x[i]);y[i]=abs(y[i]);
if(i!=1&&((x[i-1]+y[i-1])&1)!=((x[i]+y[i])&1))er();
}
m+=(!((x[n]+y[n])&1));
printf("%d\n",m);
for(RG int i=0;i<31;i++)printf("%d ",1<<i);
if(!((x[n]+y[n])&1))printf("1 ");puts("");
for(RG int i=1;i<=n;i++){
memset(k,0,sizeof(k));memset(s,0,sizeof(s));
//printf("%d,%d\n",p[i],q[i]);
if(m==32){k[32]=s[32]=1;p[i]--;q[i]--;}
for(RG int j=30;~j;j--)
if(p[i]>0)k[j+1]=1,p[i]-=(1<<j);
else k[j+1]=0,p[i]+=(1<<j);
for(RG int j=30;~j;j--)
if(q[i]>0)s[j+1]=1,q[i]-=(1<<j);
else s[j+1]=0,q[i]+=(1<<j);
for(RG int j=1;j<=m;j++)
if(k[j]&&s[j])putchar('R');
else if(k[j]&&!s[j])putchar('U');
else if(!k[j]&&s[j])putchar('D');
else if(!k[j]&&!s[j])putchar('L');
puts("");
}
return 0;
}
E-Tr/ee
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<complex>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<ctime>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define Cpy(x,y) memcpy(x,y,sizeof(x))
#define Set(x,y) memset(x,y,sizeof(x))
#define FILE "a"
#define mp make_pair
#define pb push_back
#define RG register
#define il inline
using namespace std;
typedef unsigned long long ull;
typedef vector<int>VI;
typedef long long ll;
typedef double dd;
const int N=100010;
const int mod=19940417;
const int base=26;
const dd eps=1e-6;
const int inf=2147483647;
const ll INF=1ll<<60;
const ll P=100000;
il ll read(){
RG ll data=0,w=1;RG char ch=getchar();
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
return data*w;
}
il void file(){
srand(time(NULL)+rand());
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
}
int n,m;char s[N];bool vis[N];
struct edge{int u,v;}E[N];
il void er(){puts("-1");exit(0);}
int main()
{
scanf("%s",s+1);n=strlen(s+1);if(s[n]=='1')er();
for(RG int i=1,j=n-1;i<=j;i++,j--){
if(s[i]!=s[j])er();
else if(s[i]=='0'){
if(i==1)er();
}
else{
vis[i]=1;
}
}
for(RG int i=n/2+1;i<n;i++)E[++m]=(edge){i,n};
for(RG int i=n/2,p=n;i;i--){
E[++m]=(edge){i,p};if(vis[i])p=i;
}
for(RG int i=1;i<=m;i++)printf("%d %d\n",E[i].u,E[i].v);
return 0;
}
F-Distance Sums
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<complex>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<ctime>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define Cpy(x,y) memcpy(x,y,sizeof(x))
#define Set(x,y) memset(x,y,sizeof(x))
#define FILE "a"
#define mp make_pair
#define pb push_back
#define RG register
#define il inline
using namespace std;
typedef unsigned long long ull;
typedef vector<int>VI;
typedef long long ll;
typedef double dd;
const int N=1<<17;
const int mod=998244353;
const int base=26;
const dd eps=1e-6;
const int inf=2147483647;
const ll INF=1ll<<60;
const ll P=100000;
il ll read(){
RG ll data=0,w=1;RG char ch=getchar();
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
return data*w;
}
il void file(){
srand(time(NULL)+rand());
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
}
int n,m;ll sz[N],o[N];
struct node{ll d,id;}t[N];
bool cmp_d(node a,node b){return a.d<b.d;}
bool cmp_id(node a,node b){return a.id<b.id;}
il void er(){puts("-1");exit(0);}
struct edge{int u,v;}E[N];
int head[N],nxt[N<<1],to[N<<1],cnt;
il void add(int u,int v){to[++cnt]=v;nxt[cnt]=head[u];head[u]=cnt;}
ll up[N],down[N],w[N];
void dfs1(int u,int ff){
sz[u]=1;
for(RG int i=head[u];i;i=nxt[i]){
RG int v=to[i];if(v==ff)continue;
dfs1(v,u);sz[u]+=sz[v];
up[u]+=up[v]+sz[v];
}
}
void dfs2(int u,int ff){
for(RG int i=head[u];i;i=nxt[i]){
RG int v=to[i];if(v==ff)continue;
down[v]=down[u]+up[u]-(up[v]+sz[v])+(n-sz[v]);
dfs2(v,u);
}
}
int main()
{
n=read();
for(RG int i=1;i<=n;i++){t[i].d=read();t[i].id=i;}
sort(t+1,t+n+1,cmp_d);
for(RG int i=1;i<=n;i++)o[i]=t[i].d,sz[i]=1;
for(RG int i=n,p;i!=1;i--){
p=lower_bound(o+1,o+n+1,t[i].d-n+2ll*sz[i])-o;
if(o[p]!=t[i].d-n+2ll*sz[i]||p>=i)er();
E[++m]=(edge){t[p].id,t[i].id};sz[p]+=sz[i];
}
for(RG int i=1;i<=m;i++)add(E[i].u,E[i].v),add(E[i].v,E[i].u);
dfs1(1,0);dfs2(1,0);sort(t+1,t+n+1,cmp_id);
for(RG int i=1;i<=n;i++)w[i]=down[i]+up[i];
for(RG int i=1;i<=n;i++)if(w[i]!=t[i].d)er();
for(RG int i=1;i<=m;i++)printf("%d %d\n",E[i].u,E[i].v);
return 0;
}
AtCoder Regular Contest 103 题解的更多相关文章
- AtCoder Regular Contest 127 题解
sb atcoder 提前比赛时间/fn/fn/fn--sb atcoder 还我 rating/zk/zk/zk A 签到题,枚举位数 \(+\) 前导 \(1\) 个数然后随便算算贡献即可,时间复 ...
- AtCoder Regular Contest 103 E Tr/ee
Tr/ee 思路:按照下图所示连接 代码: #pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #inclu ...
- AtCoder Regular Contest 103 Problem D Robot Arms (构造)
题目链接 Problem D 给定$n$个坐标,然后让你构造一个长度为$m$的序列, 然后给每个坐标规定一个长度为$m$的序列,ULRD中的一个,意思是走的方向, 每次从原点出发按照这个序列方向,每 ...
- AtCoder Regular Contest 103
传送门 C - /\/\/\/ 题意: 给出一个序列\(\{a_i\}\),先要求其满足以下条件: \(a_i=a_{i+2}\) 共有两个不同的数 你现在可以修改任意个数,现问最少修改个数为多少. ...
- Atcoder Regular Contest 123 题解
u1s1 我是真的不知道为什么现场这么多人切了 D,感觉 D 对思维要求显然要高于其他 300+ 人切掉的 D 吧(也有可能是 Atc 用户整体水平提升了?) A 开 幕 雷 击(这题似乎 wjz 交 ...
- AtCoder Regular Contest 094 (ARC094) CDE题解
原文链接http://www.cnblogs.com/zhouzhendong/p/8735114.html $AtCoder\ Regular\ Contest\ 094(ARC094)\ CDE$ ...
- AtCoder Regular Contest 096
AtCoder Regular Contest 096 C - Many Medians 题意: 有A,B两种匹萨和三种购买方案,买一个A,买一个B,买半个A和半个B,花费分别为a,b,c. 求买X个 ...
- AtCoder Regular Contest 061
AtCoder Regular Contest 061 C.Many Formulas 题意 给长度不超过\(10\)且由\(0\)到\(9\)数字组成的串S. 可以在两数字间放\(+\)号. 求所有 ...
- AtCoder Regular Contest 092
AtCoder Regular Contest 092 C - 2D Plane 2N Points 题意: 二维平面上给了\(2N\)个点,其中\(N\)个是\(A\)类点,\(N\)个是\(B\) ...
随机推荐
- 数据爬取后台(PHP+Python)联合作战
一. 项目声明 本项目从前端,到后台,以及分布式数据抓取,乃我一个人所写,因此项目并不太完善!在语义分析以及数据处理上并不能尽如意.但是极大的减轻了编辑的工作量! 二. 项目所用技术 本项目中前端采用 ...
- MySQL高级-查询截取分析
一.如何分析 1.观察.至少跑1天,看看生产的慢SQL情况. 2.开启慢查询日志,设置阙值比如超过5秒钟的就是慢SQL,并将它抓取出来. 3.explain + 慢SQL分析 4.show profi ...
- node环境清空控制台的代码
process.stdout.write( process.platform === 'win32' ? '\x1B[2J\x1B[0f' : '\x1B[2J\x1B[3J\x1B[H' );
- hdu1455Sticks(经典dfs+剪枝)
Sticks Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Subm ...
- cf#514B. Forgery(暴力)
B. Forgerytime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputsta ...
- Qt 5 最新信号和槽连接方式以及Lambda表达式
最近学习Qt,发现新大陆,这里做下记录. 主要内容就是原始Qt4的信号槽连接方式,以及Qt5新版的连接方式,还有件事简单演示一下lambda表达式的使用方式 代码如下 /* * 作者:张建伟 * 时间 ...
- TPO-13 C2 How to use language lab
TPO-13 C2 How to use language lab 第 1 段 1.Listen to a conversation between a student and the languag ...
- 使用flume抓取tomcat的日志文件下沉到kafka消费
Tomcat生产日志 Flume抓取日志下沉到kafka中 将写好的web项目打包成war包,eclise直接导出export,IDEA 在artifact中添加新的artifact-achieve项 ...
- 【转载】2015Android 面试题 01
1.如何避免ANR? 答:ANR:Application Not Responding,五秒在Android中,活动管理器和窗口管理器这两个系统服务负责监视应用程序的响应. 当出现下列情况时,Andr ...
- HDU 3467 Song of the Siren(圆交)
Problem Description In the unimaginable popular DotA game, a hero Naga Siren, also known as Slithice ...