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CF1322B Present

作者:互联网

题目

分析

好像是经典套路,但是我还是不会。

区间异或和并不好直接维护,于是想到按位处理。

枚举二进制的每一位,然后直接对于原序列所有数模一个 \(2^{k+1}\) (假设枚举的是 \(k\) )

然后对当前位有影响的就是和在区间 \([2^k,2^{k+1}-1]\) 和区间 \([2^k+2^{k+1},2\times (2^{k+1}-1)]\)

双指针或者二分维护这个个数即可。

代码

#include<bits/stdc++.h>
using namespace std;
//#ifdef ONLINE_JUDGE
//	#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
//	char buf[1<<21],*p1=buf,*p2=buf;
//#endif
template<typename T>
inline void read(T &x){
	x=0;bool f=false;char ch=getchar();
	while(!isdigit(ch)){f|=ch=='-';ch=getchar();}
	while(isdigit(ch)){x=x*10+(ch^48);ch=getchar();}
	x=f?-x:x;
	return ;
}
template<typename T>
inline void write(T x){
	if(x<0) x=-x,putchar('-');
	if(x>9) write(x/10);
	putchar(x%10^48);
	return ;
}
#define ll long long
#define ull unsigned long long
#define ld long double
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pc putchar
#define PII pair<int,int>
#define rep(i,x,y) for(register int i=(x);i<=(y);i++)
#define dep(i,y,x) for(register int i=(y);i>=(x);i--)
#define repg(i,x) for(int i=head[x];i;i=nex[i])
#define filp(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
#define infilp(s) freopen(s".in","r",stdin)
#define outfilp(s) freopen(s".out","w",stdout)
const int MOD=1e9+7;
inline int inc(int x,int y){x+=y;return x>=MOD?x-MOD:x;}
inline int dec(int x,int y){x-=y;return x<0?x+MOD:x;}
inline void incc(int &x,int y){x+=y;if(x>=MOD) x-=MOD;}
inline void decc(int &x,int y){x-=y;if(x<0) x+=MOD;}
inline void chkmin(int &x,int y){if(y<x) x=y;}
inline void chkmax(int &x,int y){if(y>x) x=y;}
const int N=4e5+5,M=2e5+5,INF=1e9+7;
int n,m,a[N],b[N],Ans;
signed main(){
//	double ST=clock();
	// ios::sync_with_stdio(false);
//#ifndef ONLINE_JUDGE
//	filp("my");
//#endif
	read(n);
	for(int i=1;i<=n;i++) read(a[i]);
	for(int k=0;k<=28;k++){
		const int now=(1<<k);
		int res=0;
//		cout<<k<<":";
		for(int i=1;i<=n;i++) b[i]=a[i]&((now<<1)-1);
//		cout<<endl;
		sort(b+1,b+n+1);
		for(int i=1;i<=n;i++){
			int pos1=lower_bound(b+1,b+n+1,now-b[i])-b,
				pos2=upper_bound(b+1,b+n+1,(now<<1)-1-b[i])-b,
				pos3=lower_bound(b+1,b+n+1,(now<<1)+now-b[i])-b;
//			pos2--;
			res+=(n-pos1+1)-(n-pos2+1)+(n-pos3+1);
			if(pos1<=i) res--;
			if(pos2<=i) res++;
			if(pos3<=i) res--;
		}
		res/=2;
//		cout<<k<<":"<<res<<endl;
		if(res&1) Ans+=now;
	}
	write(Ans);
//	cerr<<"\nTime:"<<(clock()-ST)/CLOCKS_PER_SEC<<"s\n";
	return 0;
}
/*

*/

标签:ch,int,long,CF1322B,inline,define,Present,MOD
来源: https://www.cnblogs.com/Akmaey/p/15617932.html