201,600-degree star Supplement preface

2016 "Baidu Star"-Rematch (Astar Round3)

Take Pictures

Idea: first project all the line segments to the x-axis, and then put the end coordinates of all the segments into an array, sorted and enumerated from the smallest start of the coordinates. If a start sign is encountered, add one; end point sign minus one. This allows you to find out how many segments are included in the same moment when you encounter the end point of the current segment

/************************************************************** problem:hdu 5417 user:youmi language:c++ R esult:accepted time:1591ms memory:1816k****************************************************************///#pragma COMMENT (linker, "/stack:1024000000,1024000000")//#include <bits/stdc++.h>#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<stack>#include<Set>#include<sstream>#include<cmath>#include<queue>#include<deque>#include<string>#include<vector>#defineZeros (a) memset (A,0,sizeof (a))#defineOnes (a) memset (A,-1,sizeof (a))#defineSC (a) scanf ("%d", &a)#defineSC2 (A, b) scanf ("%d%d", &a,&b)#defineSC3 (a,b,c) scanf ("%d%d%d", &a,&b,&c)#defineSCS (a) scanf ("%s", a)#defineSclld (a) scanf ("%i64d", &a)#definePT (a) printf ("%d\n", a)#definePtlld (a) printf ("%i64d\n", a)#defineRep (i,from,to) for (int i=from;i<=to;i++)#defineIrep (I,to,from) for (int i=to;i>=from;i--)#defineMax (a) (a) > (b)? ( A):(B))#defineMin (a) < (b) ( A):(B))#defineLson (step<<1)#defineRson (lson+1)#defineEPS 1e-6#defineOO 0x3fffffff#defineTEST cout<< "*************************" <<endlConst DoublePi=4*atan (1.0);using namespaceStd;typedefLong LongLl;inlineintRead () {intx=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9') {if(ch=='-') f=-1; Ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; Ch=GetChar ();} returnx*F;}Const intmaxn=20000+Ten;intN,cnt,x,y,z,d,po,lmax,l,r,ans;structdata{intx, Y, Z;} A[MAXN];BOOLCMP (data a,data b) {if(a.x!=b.x)returna.x<b.x; Else    {        if(A.Y!=B.Y)returna.y<b.y;Else returna.z<b.z; }}intMain () {#ifndef Online_judge freopen ("In.txt","R", stdin); #endif    intt_t; scanf ("%d",&t_t);  for(intKase=1; kase<=t_t;kase++) {n=read (); Cnt=0;  for(intI=1; i<=n;i++) {x=read (), Y=read (), Z=read (), d=read (); if(y-x<=2*z) {if(d==1) po=0;Elsepo=1; a[++cnt].x=y-z; a[cnt].y=0; a[cnt].z=PO; a[++cnt].x=x+z; a[cnt].y=1; a[cnt].z=PO; }} sort (a+1, a+cnt+1, CMP); Lmax=l=r=ans=0;  for(intI=1; i<=cnt;i++)        {            if(a[i].z==0)            {                if(a[i].y==0) ++l;Else--l; }            Else            {                if(a[i].y==0) ++r;Else--R; }            if(L>lmax) lmax=l; if(Lmax+r>ans) ans=lmax+R; } printf ("Case #%d:\n%d\n", Kase,ans); }    return 0;}

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2016 "Baidu Star"-Preliminary round (Astar round2a)

1001 All X

Idea: x* (10n-1)/9, after the should be simple

1002 Sitting in line

This has been written in the blog, so the title has been connected to the written blog.

1005 BD String

Ideas:

1---B

BBD 2---

3---bbd| B| Bdd

4---bbd| B| bdd| B| bbd| D| Bdd

5---bbd| B| bdd| B| bbd| D| bdd| B| bbd| B| bdd| D| bbd| D| Bdd

From the fifth line can see the law: every 8 for a group, wherein BBD,BDD continuous repetition, 4 in multiples of the corresponding letter, but also the occurrence of the law, then this part of the DFS can be; then for parts less than or equal to 8, you can use an array preprocessing.

Note: (Parts less than or equal to 8) and (multiples of 4) are calculated once at position 4, so remove this part

/************************************************************** problem:hdu 5694 user:youmi language:c++ R esult:accepted time:0ms memory:1572k****************************************************************///#pragma COMMENT (linker, "/stack:1024000000,1024000000")//#include <bits/stdc++.h>#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<stack>#include<Set>#include<sstream>#include<cmath>#include<queue>#include<deque>#include<string>#include<vector>#defineZeros (a) memset (A,0,sizeof (a))#defineOnes (a) memset (A,-1,sizeof (a))#defineSC (a) scanf ("%d", &a)#defineSC2 (A, b) scanf ("%d%d", &a,&b)#defineSC3 (a,b,c) scanf ("%d%d%d", &a,&b,&c)#defineSCS (a) scanf ("%s", a)#defineSclld (a) scanf ("%i64d", &a)#definePT (a) printf ("%d\n", a)#definePtlld (a) printf ("%i64d\n", a)#defineRep (i,from,to) for (int i=from;i<=to;i++)#defineIrep (I,to,from) for (int i=to;i>=from;i--)#defineMax (a) (a) > (b)? ( A):(B))#defineMin (a) < (b) ( A):(B))#defineLson (step<<1)#defineRson (lson+1)#defineEPS 1e-6#defineOO 0x3fffffff#defineTEST cout<< "*************************" <<endlConst DoublePi=4*atan (1.0);using namespaceStd;typedefLong Longll;inline ll Read () {ll x=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9') {if(ch=='-') f=-1; Ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; Ch=GetChar ();} returnx*F;}intb[]={0,1,2,2,0,3,3,3,0};ll Sovle (ll tt) {if(tt==0)        return 0; ll ans=0; Ans+=sovle (tt/4); if(tt<4)        returnB[TT]; Else if(tt<8)    {        if(tt==4)            returnans+b[3]; Else            returnans+B[TT]; } ans+=3* (tt/8); Ans+=sovle (tt%8); if(tt%8>=4) ans-=1; returnans;}intMain () {#ifndef Online_judge freopen ("In.txt","R", stdin); #endif    intt_t; scanf ("%d",&t_t);  for(intKase=1; kase<=t_t;kase++) {ll l,r; L=read (), r=read (); ll ans=Sovle (R); Ans-=sovle (l1);    Ptlld (ANS); }}

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1006 Gym Class

This is also in the 01 dynamic planning of the first question has been summed up, so here also directly link to the corresponding blog

201,600-degree star Supplement preface