4245: [Ontak2015]or-xor

4245: [Ontak2015]or-xor

https://www.lydsy.com/JudgeOnline/problem.php?id=4245

1 /*2 The requirements are divided into m parts, the total value is a1|a2|a3 ..., the total value is minimal, AI is the XOR of the first part. 3  4 The first preprocessing is XOR. 5 According to the greedy thought, the high position is preferably 0. 6 so from high to low enumeration, see if each bit can be 0. 7 if the current bit can be 0, each copy (XOR) (or) is 0, so each share of the XOR or the bit cannot have 18 then a point can become a split point when and only if this bit of the sum value of the point is 0, then count how many points the sum is 09 If the number is >=m and all 1 on this one is an even number (sum[n]& (1ll<<i) ==0), then this bit can be 0, and then for those points that cannot be split, the marker can no longer become a split point. Ten Otherwise it would make this person not 0.  One if not 0, that would be 1, ans update.  A */ -#include <bits/stdc++.h> - using namespacestd; thetypedefLong LongLL; -   - inline LL Read () { -LL x=0, f=1;CharCh=getchar (); for(;! IsDigit (CH); Ch=getchar ())if(ch=='-') f=-1; +      for(; isdigit (ch); Ch=getchar ()) x=x*Ten+ch-'0';returnx*F; - } +   A Const intN =500100; at LL A[n],sum[n],flag[n]; -   - intMain () { -     intN,m;cin >> N >>m; -      for(intI=1; i<=n; ++i) -A[i] = Read (), sum[i] = sum[i-1] ^A[i]; inLL ans =0;  -      for(intI= +; i>=0; -I.) {//See if I can be 0 for the first place . to         intCNT =0; +          for(intj=1; j<=n; ++J) {//count how many right endpoints meet the criteria -             if(!flag[j] && (sum[j]& (1ll<<i)) = =0) cnt++; the         } *         if(CNT >= m && (sum[n]& (1ll<<i)) = =0) {//If this person can put 0. ----Less on curly braces ...  $              for(intj=1; j<=n; ++J)//Mark All points that cannot be the right endpointPanax Notoginseng                 if((sum[j]& (1ll<<i))! =0) Flag[j] =1; -         } the         ElseAns |= (1LL <<i); +     } Acout <<ans; the     return 0; +}

4245: [Ontak2015]or-xor