[SCOI 2016] 美味

[题目链接]

          https://www.lydsy.com/JudgeOnline/problem.php?id=4571

[算法]

        二分 + 可持久化线段树逐位确定答案的每一位即可

        时间复杂度 : O(NlogN^2)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int MAXLOG = 30;
const int MAXP = N * MAXLOG;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

int n , m;
int a[N] , rt[N];

struct Presitent_Segment_Tree
{
        int sz;
        int latest[MAXP] , lc[MAXP] , rc[MAXP];
        inline void init()
        {
                sz = 0;
        }
        inline void modify(int &now , int old , int l , int r , int x , int loc)
        {
                now = ++sz;
                lc[now] = lc[old] , rc[now] = rc[old];
                latest[now] = loc;
                if (l == r) return;
                int mid = (l + r) >> 1;
                if (mid >= x) modify(lc[now] , lc[old] , l , mid , x , loc);
                else modify(rc[now] , rc[old] , mid + 1 , r , x , loc);
        }
        inline bool query(int now , int lft , int l , int r , int ql , int qr)
        {
                if (ql < 0) 
                        return false;
                if (ql > qr)
                        return false;
                if (l == ql && r == qr)
                        return latest[now] >= lft;
                int mid = (l + r) >> 1;
                if (mid >= qr) return query(lc[now] , lft , l , mid , ql , qr);
                else if (mid + 1 <= ql) return query(rc[now] , lft , mid + 1 , r , ql , qr);
                else return query(lc[now] , lft , l , mid , ql , mid) | query(rc[now] , lft , mid + 1 , r , mid + 1 , qr);
        }
} PST;

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}

int main()
{
        
        read(n); read(m);
        PST.init();
        for (int i = 1; i <= n; i++)
        {
                read(a[i]);
                PST.modify(rt[i] , rt[i - 1] , 0 , 300000 , a[i] , i);
        }
        for (int i = 1; i <= m; i++)
        {
                int b , x , l , r , ans = 0;
                read(b); read(x); read(l); read(r);
                for (int j = 17; j >= 0; j--)
                {
                        int now = ans + ((((b >> j) & 1) ^ 1) << j);
                        if (PST.query(rt[r] , l , 0 , 300000 , max(0 , now - x) , now + (1 << j) - 1 - x)) ans = now;
                        else ans = ans + (((b >> j) & 1) << j);
                }    
                printf("%d\n" , ans ^ b);
        }
        
        return 0;
    
}