Uvalive 3177 Great Wall Guard

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&itemid=8&page=show_problem& problem=1178

Http://7xjob4.com1.z0.glb.clouddn.com/8e1ee1ef5ea10e46ebda75b88b058f47

Test instructions: N personal circle, each person has a different or equal number of gifts, adjacent gifts can not be the same, the minimum total number of gifts.

Idea: N is even: the maximum number of gifts for two adjacent people; n is an odd number: two-way choice of gift quantity, see if it is feasible, set the first person gift for the former, the number of even-numbered people try to take in front, odd people try to take in the back, and finally check whether the last person's gift and the first person

1#include <bits/stdc++.h>2 using namespacestd;3 4 intN;5 intr[100005],left[100005],right[100005];6 7 intCheckintp)8 {9     inti,j;Ten     intx=r[1],y=p-r[1]; Oneleft[1]=x,right[1]=0; A      for(i=2; i<=n;i++) -     { -         if(i%2==0) the         { -Left[i]=min (x-left[i-1],r[i]); -right[i]=r[i]-Left[i]; -         } +         Else -         { +Right[i]=min (y-right[i-1],r[i]); Aleft[i]=r[i]-Right[i]; at         } -     } -     if(left[n]==0) -         return 1; -     return 0; - } in  - intMain () to { +     inti,j; -      while(SCANF ("%d", &n)!=eof && n!=0) the     { *          for(i=1; i<=n;i++) $         {Panax Notoginsengscanf"%d",&r[i]); -         } the         if(n==1) +         { Aprintf"%d\n", r[1]); the             Continue; +         } -  $         intL=0, r=0; r[n+1]=r[1]; $          for(i=1; i<=n;i++) L=max (l,r[i]+r[i+1]); -         if(n%2==1) -         { the              for(i=1; i<=n;i++) R=max (r,r[i]*3); -              while(l<R)Wuyi             { the                 intMid= (L+R)/2; -                 if(check (mid)) Wu                 { -R=mid; About                 } $                 Else -                 { -L=mid+1; -                 } A             } +         } theprintf"%d\n", L); -     } $     return 0; the}

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Uvalive 3177 Great Wall Guard