Comet OJ

题目描述

求(x,y)的对数满足x∈[0,a],y∈[0,b],x⊕y=0且|x-y|<=m

题解

一种比较sb的做法是考虑x-y的借位，根据借位以及差值进行转移

还有一种比较正常的做法，假设一开始x=0,y=n，那么就需要把y的某一些1移到x上，也就是对于(x-y)加上2^(i+1)

设加的数之和为s，那么需要保证|s-n|<=m，也就是n-m<=s<=n+m

注意是当n的第i位为1时才可以加上2^(i+1)，把上界右移一位后就变成加上2^i

设f[i][0/1][0/1][0/1]，表示当前到第i位，x、y、s是否顶住上界

枚举第i位的xy所选，使得x[i]^y[i]=n[i]且新的s不会越界即可

code

sb版

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define min(a,b) (a<b?a:b)
using namespace std;

int a[60];
int b[60];
int x[60];
int y[60];
long long f[61][2][2][2][2];
int T,i,j,k,l,I,J,K,L,p,q,P,Q,fs,ll;
long long n,m,A,B,ans;

int swap(int &x,int &y)
{
	int z=x;
	x=y;
	y=z;
}

void work()
{
	memset(f,0,sizeof(f));
	f[60][0][0][0][0]=1;
	fd(i,60,1)
	{
		I=i-1;
		
		fo(j,0,1)
		{
			fo(k,0,1)
			{
				fo(p,0,1)
				{
					fo(q,0,1)
					if (f[i][j][k][p][q])
					{
						if (!a[I])
						{
							if (j && !b[I]) continue;
							
							fo(l,0,1)
							if ((p || x[I]>=l) && (q || y[I]>=l))
							{
								if (!j && b[I]>0) K=1; else K=k;
								if (x[I]>l) P=1; else P=p;
								if (y[I]>l) Q=1; else Q=q;
								
								f[I][j][K][P][Q]+=f[i][j][k][p][q];
							}
						}
						else
						{
							if (I==fs) ll=1;
							else ll=0;
							
							fo(l,ll,1)
							if ((p || x[I]>=l) && (q || y[I]>=(1-l)))
							{
								if (!l)
								{
									if (!j)
									J=0,K=1;
									else
									{
										if (b[I])
										J=0,K=k;
										else
										J=1,K=k;
									}
								}
								else
								{
									if (j && !b[I]) continue;
									
									if (k)
									J=0,K=1;
									else
									{
										if (!b[I])
										{
											if (j)
											continue;
											else
											J=1,K=0;
										}
										else
										{
											if (j)
											continue;
											else
											J=0,K=0;
										}
									}
								}
								if (x[I]>l) P=1; else P=p;
								if (y[I]>(1-l)) Q=1; else Q=q;
								
								f[I][J][K][P][Q]+=f[i][j][k][p][q];
							}
						}
					}
				}
			}
		}
	}
	
	fo(k,0,1)
	{
		fo(p,0,1)
		{
			fo(q,0,1)
			ans+=f[0][0][k][p][q];
		}
	}
}

int main()
{
	scanf("%d",&T);
	for (;T;--T)
	{
		scanf("%lld%lld%lld%lld",&A,&B,&n,&m);
		
		fs=-1;
		
		fo(i,0,59)
		{
			a[i]=n&1;
			n>>=1;
			
			if (a[i])
			fs=i;
		}
		
		if (fs==-1)
		{
			printf("%lld
",min(A,B)+1);
			continue;
		}
		
		fo(i,0,59) b[i]=m&1,m>>=1;
		fo(i,0,59) x[i]=A&1,A>>=1;
		fo(i,0,59) y[i]=B&1,B>>=1;
		
		ans=0;
		work();
		
		fo(i,0,59)
		swap(x[i],y[i]);
		
		work();
		
		printf("%lld
",ans);
	}
}

正常版

#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
using namespace std;

int a[61];
int x[61];
int y[61];
int z[61];
long long f[61][2][2][2];
int T,i,j,k,l,I,J,K,L,s,S,X,Y,X2,Y2;
long long N,M,A,B,s1;

void turn(int *a,long long s)
{
	int i;
	
	fo(i,1,60)
	a[i]=s%2,s/=2;
}

long long work(long long t)
{
	long long ans=0;
	
	if (t>=0)
	{
		t/=2;
		turn(a,t);
	}
	else
	return 0;
	
	memset(f,0,sizeof(f));
	f[60][0][0][0]=1;
	
	fd(i,60,1)
	{
		I=i-1;
		
		fo(j,0,1)
		{
			fo(k,0,1)
			{
				fo(l,0,1)
				if (f[i][j][k][l])
				{
					if (j) X2=1; else X2=x[i];
					if (k) Y2=1; else Y2=y[i];
					
					fo(X,0,X2)
					{
						if (X<x[i]) J=1; else J=j;
						
						fo(Y,0,Y2)
						if ((X^Y)==z[i] && !(!l && X && !Y && !a[i]))
						{
							if (Y<y[i]) K=1; else K=k;
							if ((X && !Y)<a[i]) L=1; else L=l;
							
							f[I][J][K][L]+=f[i][j][k][l];
						}
					}
				}
			}
		}
	}
	
	fo(j,0,1)
	{
		fo(k,0,1)
		{
			fo(l,0,1)
			ans+=f[0][j][k][l];
		}
	}
	
	return ans;
}

int main()
{
	scanf("%d",&T);
	for (;T;--T)
	{
		scanf("%lld%lld%lld%lld",&A,&B,&N,&M);
		turn(x,A);
		turn(y,B);
		turn(z,N);
		
		printf("%lld
",work(N+M)-work(N-M-1));
	}
}