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\n",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\n",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\n",work(N+M)-work(N-M-1));
}
}