BestCoder Round #79 (div.2) 1002 Claris and XOR

来源：互联网发布：sql 注入式攻击编辑：程序博客网时间：2024/05/29 13:59

题意很简洁，给出两个区间[a,b],[c,d]且a,b,c,d均不超过1018

一般求最大异或值得解法就是贪心，从高位往低位贪即可，字典树也是这个原理，只不过上者需要O(n)的预处理，这题关键还是要知道若到了某一位两个数0,1均能取，那么后面肯定全为1。这样只要扫一遍就可以了。

#include <iostream>#include <cstdio>#include <map>#include <queue>#include <vector>#include <climits>#include <cstring>#include <cmath>#include <algorithm>#define LL long longusing namespace std;LL a, b, c, d;bool judge(LL l, LL r, LL n, LL m){    if(l < n)    {        if(r >= n)return true;    }    if(l >= n && l <= m)return true;    return false;}int main(){    int T;    scanf("%d", &T);    while(T--)    {        scanf("%I64d%I64d%I64d%I64d", &a, &b, &c, &d);        LL l, r, left , right, ans;        l = r = 0LL;        bool la, lb, ra, rb;        bool flag = false;        for(LL i = 60LL; i >= 0LL; --i)        {            ra = rb = la = lb = false;            left = l;            right = l + (1LL << i) - 1LL;            if(judge(left, right, a, b))la = true;            left = l + (1LL << i);            right = l + (1LL << (i + 1LL)) - 1LL;            if(judge(left, right, a, b))lb = true;            left = r;            right = r + (1LL << i) - 1LL;            if(judge(left, right, c, d))ra = true;            left = r + (1LL << i);            right = r + (1LL << (i + 1LL)) - 1LL;            if(judge(left, right, c, d))rb = true;            if(ra && rb && la && lb)            {                ans = (l ^ r) + (1LL << (i + 1LL)) - 1LL;                flag = true;                break;            }            if(ra && lb)            {                l = l + (1LL << i);                continue;            }            if(rb && la)            {                r = r + (1LL << i);                continue;            }            if(ra && la)            {                continue;            }            if(rb && lb)            {                l = l + (1LL << i);                r = r + (1LL << i);            }        }        if(!flag) ans = l ^ r;        cout << ans << endl;    }    return 0;}

0 0