Uva Great Wall guard--1335-beijing guards

Two-point search + some tricks

1#include <iostream>2 using namespacestd;3 4 Const intmaxn=100000+Ten;5 intn,r[maxn],left[maxn],right[maxn];//because no calculation scheme, so you can press "1~r[i" and "r[i]+1~p" each take a few points, then did not think of this with set class write a time-consuming ~ ~ ~ (>_ <) ~ ~ ~6 7 BOOLOkintp)8 {9     intx=r[1],y=p-r[1];Tenleft[1]=x; right[1]=0; One      for(intI=2; i<=n;i++) A     { -         if(i&1) -         { theRight[i]=min (y-right[i-1],r[i]);//Odd guard first take the right. -left[i]=r[i]-right[i];//Look at the left to a few -         } -         Else +         { -Left[i]=min (x-left[i-1],r[i]);//Even the guard first take the left. +right[i]=r[i]-left[i];//look at the right to a few A         } at     } -     returnleft[n]==0;//The Nth guard on the left is 0 o'clock - } -  - intMain () - { in      while(cin>>n&&N) -     { to          for(intI=1; i<=n;i++) +Cin>>R[i]; -         if(n==1) {cout<<r[1]<<endl;Continue;} This place pits me. ther[n+1]=r[1]; *         intL=0, r=0; $          for(intI=1; i<=n;i++) L=max (l,r[i]+r[i+1]);Panax Notoginseng         if(n&1) -         { the              for(intI=1; i<=n;i++) R=max (r,r[i]*3); +              while(l<R) A             { the                 intM= (L+R)/2; +                 if(OK (M)) R=M; -                 Elsel=m+1; $             } $         } -cout<<l<<Endl; -     } the}

Uva Great Wall guard--1335-beijing guards