hdu 6889 数位dp Xor

Posted 2021-10-10 goto_1600

题意：

思路：
数位dp，考虑设计状态，第四个条件很好满足，设计一个flag3为状态位就可以了代表是否已经大于w，flag2,flag3代表是否到达枚举x和y的limit，那么第三个条件怎么设计呢，考虑稍微改变一下柿子，$x-y<=k\&\&y-x<=k$那么就好办了，考虑x-y+k只有$[-1,2]$的情况下，由于前面如果推下来<-1那么永远都不能大于，是无效状态，同时2这个状态位也是多余的，有1就够了，因为后面怎么弄都不会超过这个1.

//#pragma GCC target("avx")
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast")
// created by myq
#include<iostream>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cctype>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<set>
#include<map>
#include<sstream>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
#define x first
#define y second
typedef pair<int,int> pii;
const int N = 400010;
const int mod=998244353;
inline int read()
{
	int res=0;
	int f=1;
	char c=getchar();
	while(c>'9' ||c<'0')
	{
		if(c=='-')	f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9')
	{
		res=(res<<3)+(res<<1)+c-'0';
	}
	return res;
 }
ll dp[32][2][2][2][4][4];
int a[32],a1[32];
int b[32],a2[32];
int c[32],k[32];
int d[32],w[32];
int E=1;
ll dfs(int len,int flag1,int flag2,int flag3,int k1,int k2){
	
	if(!len){
		return k1>=0 && k2>=0;
	}
	if(~dp[len][flag1][flag2][flag3][k1+E][k2+E])	return dp[len][flag1][flag2][flag3][k1+E][k2+E];
	ll res=0;
	int up1=flag1?a[len]:1;
	int up2=flag2?b[len]:1;
	for(int i=0;i<=up1;i++)
	{
		for(int j=0;j<=up2;j++){
			if(flag3 && (i^j)&&!w[len])	continue;
			int kk1=min(k1*2+k[len]+j-i,2);
			int kk2=min(k2*2+k[len]+i-j,2);
			if(kk1<=-2 || kk2<=-2)	continue;
			res+=dfs(len-1,flag1&(i==up1),flag2&(j==up2),flag3&((i^j)==w[len]),kk1,kk2);
		}
	}
	
	return dp[len][flag1][flag2][flag3][k1+E][k2+E]=res;
	
}
ll solve(int A,int B,int K,int W){
	memset(dp,-1,sizeof dp);
	for(int i=30;i>=0;i--){
		a[i+1]=A>>i&1; 
		b[i+1]=B>>i&1;
		k[i+1]=K>>i&1;
		w[i+1]=W>>i&1;
	}
	return dfs(31,1,1,1,0,0);
}
int main()
{
	int T;
	cin>>T;
	while(T--){
		int a,b,k,w;
		cin>>a>>b>>k>>w;
		cout<<solve(a,b,k,w)<<endl;
	}
	return 0;

}
/**
* In every life we have some trouble
* When you worry you make it double
* Don't worry,be happy.
**/