HDU 5661 Claris and XOR(xor贪心)

题意:

$给定a,b,c,d,1\leq a,b,c,d\leq10^{18}$
$现要求找到x\oplus y,x\in [a, b],y\in[c, d]的最大值$

分析:

$显然从高位到低位开始贪心,要最大肯定优先要求不同$
$也就是先测试01和10这两种情况,再测试00和11这两种情况,同时判断是不是符合区间范围$
$假设i这位已经放置好了,显然最小值后面全是0,最大值即全是1$
$考虑合法比较麻烦,反向思考非法情况,最小值比右边界大,或最大值比左边界小$
$时间复杂度O(b),b=log_2 10^{18}$

代码:

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//
// Created by TaoSama on 2016-04-25
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
typedef long long LL;
LL a, b, c, d;
bool ok(LL x, LL y, int i) {
LL lft = (1LL << i) - 1;
LL xr = x + lft, yr = y + lft;
if(x > b || xr < a) return false;
if(y > d || yr < c) return false;
return true;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
while(t--) {
scanf("%I64d%I64d%I64d%I64d", &a, &b, &c, &d);
LL x = 0, y = 0;
for(int i = 62; ~i; --i) {
LL delta = 1LL << i;
if(ok(x + delta, y, i)) x += delta;
else if(ok(x, y + delta, i)) y += delta;
else if(ok(x + delta, y + delta, i)) {
x += delta;
y += delta;
}
}
printf("%I64d\n", x ^ y);
}
return 0;
}


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