noip模拟赛 区间

分析：要遍历所有的区间，肯定是枚举左端点，然后再枚举右端点.关键是怎么高效地求区间&,|,一般而言是用前缀和的，但是&,|不满足区间可减性，所以可以考虑线段树？这道题不带修改操作，用线段树太浪费了，那么可以用ST表来维护.

查询做到O(1)了，但是怎么快速枚举区间呢？枚举左端点和右端点肯定只能选择一个优化，优化枚举右端点的循环.观察数据范围,100000,很容易想到二分.可以每次固定左端点，然后二分右端点的位置.因为&操作随着区间数的增加而答案减少，|是增加，都满足单调性，所以求满足两个条件的区间的交，统计一下区间的元素个数有多少个就可以了.

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
const long long inf = 1LL << , mod = 1e9 + ;

typedef long long ll;

ll n, a, b, c, d, f1[][], f2[][], s[], ans;

void init()
{
    for (int j = ; j <= ; j++)
        for (int i = ; i + ( << j) -  <= n; i++)
        {
            f1[i][j] = f1[i][j - ] & f1[i + ( << (j - ))][j - ];
            f2[i][j] = f2[i][j - ] | f2[i + ( << (j - ))][j - ];
        }
}

ll query(ll l, ll r,ll op)
{
    ll k = (ll)((log(r - l + )) / log(2.0));
    if (op == )
        return f1[l][k] & f1[r - ( << k) + ][k];
    else
        return f2[l][k] | f2[r - ( << k) + ][k];
}

int main()
{
    scanf("%lld%lld%lld%lld%lld", &n, &a, &b, &c, &d);
    for (int i = ; i <= n; i++)
    {
        scanf("%lld", &s[i]);
        f1[i][] = f2[i][] = s[i];
    }
    init();
    for (int i = ; i <= n; i++)
    {
        ll l = i, r = n, temp1 = inf, temp2 = -inf, temp3 = inf, temp4 = -inf;
        while (l <= r)
        {
            ll mid = (l + r) >> ;
            if (query(i, mid,) >= a)
            {
                l = mid + ;
                temp2 = mid;
            }
            else
                r = mid - ;
        }

        l = i, r = n;
        while (l <= r)
        {
            ll mid = (l + r) >> ;
            if (query(i, mid, ) <= b)
            {
                r = mid - ;
                temp1 = mid;
            }
            else
                l = mid + ;
        }

        l = i, r = n;
        while (l <= r)
        {
            ll mid = (l + r) >> ;
            if (query(i, mid, ) >= c)
            {
                r = mid - ;
                temp3 = mid;
            }
            else
                l = mid + ;
        }

        l = i, r = n;
        while (l <= r)
        {
            ll mid = (l + r) >> ;
            if (query(i, mid, ) <= d)
            {
                l = mid + ;
                temp4 = mid;
            }
            else
                r = mid - ;
        }
        ll ll = max(temp1, temp3), rr = min(temp2, temp4);
        ans += max((long long), rr - ll + );
        ans %= mod;
    }
    printf("%lld\n", ans);

    return ;
}