链接：https://ac.nowcoder.com/acm/contest/897/L
来源：牛客网

XOR
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
Exclusive or is a logical operation that outputs true only when inputs differ(one is true, the other is false). It is symbolized by the infix operators such as XOR,
⊕
⊕.
This time, brave QQQ raises a problem to you. Given an interval [l, r], you need to calculate how many numbers x between l and r, where x satisfies
x
⊕
4
x
⊕
5
x
=
0
x⊕4x⊕5x=0.
输入描述:
The first line contains an integer number T, the number of test cases.

i
t
h
ith of each next T lines contains two integers l, r(
1
≤
l
≤
r
≤
10
18
1≤l≤r≤1018).
输出描述:
For each test case print the answer.
示例1
输入
复制
2
2 6
1 109
输出
复制
4
39

题意：

思路：
根据异或的规律，我们知道x^x=0, ^是异或运算，即两个相等的数异或起来为0，

又因为异或运算满足交换律和分配律。

所以 x⊕4x⊕5x=0. 可以得到，(x⊕4x)⊕5x=0.

那么当x⊕4x=5x 使满足条件，我们还知道x⊕4x=x+4x 当且仅当 x与4x 在二进制状态下，任一位不同时为1.

而4*x 就是x在二进制状态下 尾部补两个0，也就是左移2位，那么为了满足上面的条件也就要满足 二进制数中没有两个1中间只有一个数。

例如二进制中不能有 101 ，111 ，这种子串。

这显然就是数位dp了，直接爆搜肯定过不去，加一个记忆化搜索即可。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2) { ans = ans * a % MOD; } a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/int a[700];
int cnt;
ll dp[70][2][2];
ll dfs(int dep, int last1, int last2, bool limit)
{ll res = 0ll;if (dep == 0) {return 1ll;} else {if (limit) {int up = a[dep];for (int i = 0; i <= up; ++i) {if (i) {if (last2!=1) {res += dfs(dep - 1, i, last1, limit && (i == a[dep]));}} else {res += dfs(dep - 1, i, last1, limit && (i == a[dep]));}}return res;} else {if(dp[dep][last1][last2]!=-1){return dp[dep][last1][last2];}int up = 1;for (int i = 0; i <= up; ++i) {if (i) {if (last2!=1){res += dfs(dep - 1, i, last1, limit && (i == a[dep]));}} else {res += dfs(dep - 1, i, last1, limit && (i == a[dep]));}}dp[dep][last1][last2]=res;return res;}}}ll solve(ll x)
{cnt = 0;while (x) {if (x & 1) {a[++cnt] = 1;} else {a[++cnt] = 0;}x >>= 1;}return dfs(cnt, 0, 0, 1);
}
int main()
{//freopen("D:\\code\\text\\input.txt","r",stdin);//freopen("D:\\code\\text\\output.txt","w",stdout);int t;gbtb;cin >> t;ll l, r;memset(dp,-1,sizeof(dp));while (t--) {cin >> l >> r;cout << solve(r) - solve(l - 1) << endl;}return 0;
}inline void getInt(int *p)
{char ch;do {ch = getchar();} while (ch == ' ' || ch == '\n');if (ch == '-') {*p = -(getchar() - '0');while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 - ch + '0';}} else {*p = ch - '0';while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 + ch - '0';}}
}