cf1058e thinking prefix processing

Source: Internet
Author: User

The question is: given a length of(n≤3x105) of the seriesai (1≤< Span id= "mathjax-span-21" class= "Msubsup" >a i≤ 1018) Exchange a number of arbitrary bits, ask how many intervals you can select after Operation XOR 0

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Idea: Because of the binary exchange at will, then its own value does not have to consider, just want to save it has how many binary 1 of the number is good.

Sufficient and necessary conditions:

    • The number of binary 1 in the interval is even
    • The number of bits in the interval is less than or equal to half of the number of the binary

For condition One

So the parity of (L,R) is judged by the parity of (0,l), so that the parity of all the prefixes is counted, and the degree of responsibility of O (n) is calculated.

For the time being, add all the conditions and subtract from the condition two.

Because AI <1e18 the >1 limit, enumerating about 60 bits can be as

#include <bits/stdc++.h>using namespacestd;#definell Long Long#definePB Push_back#defineFi first#defineSe Second#defineAll (v) v.begin (), V.end ()#defineForn (i,a,n) for (int i=a;i<n;++i)Const intN = 3e5+4; ll A[n],b[n];ll Cbit (ll x) {ll res=1;  while(x) {if(x&1) res++; X/=2; }    returnres-1;} ll Sum[n];intcnt[2];intMain () {intN; CIN>>N; Forn (i,1, n+1) {scanf ("%lld", A +i); B[i]=cbit (A[i]); } ll ans=0 ; cnt[0]=1; //1 (1..j-1), J is even//here 's a formula to illustrateForn (I,1, n+1) {Sum[i]= sum[i-1]+B[i]; Ans+ = cnt[sum[i]&1];//1 (0..J), J is even//here 's a formula to illustrate//In accordance with condition 2, minus the past.        intJ =i,k=i; ll Max=B[i];  while(k>=1&& j-k<= +) {Max=Max (max,b[k]); if(max*2> sum[i]-sum[k-1] && ((sum[i]-sum[k-1] )%2==0)) ans--; K--; } Cnt[sum[i]&1]++; //cout<<ans<<endl;} cout<<ans<<Endl; return 0;}

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cf1058e thinking prefix processing

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