Hdu4737 ( A Bit Fun ) 线段树

题意：统计最后有多少对[i,j]使得其区间内所有的值的或的值<m

| 是非递减运算，线段树维护区间和 然后顺序统计下。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include <math.h>
using namespace std;
typedef long long LL;

const LL maxn = 111111;
LL sum[maxn<<2];
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
void up(LL rt)
{
    sum[rt] = sum[rt << 1] | sum[rt << 1 | 1];
}

void build(LL l, LL r, LL rt)
{
    if (l == r){
        scanf("%I64d", &sum[rt]); return;
    }
    LL mid = (l + r) >> 1;
    build(lson);
    build(rson);
    up(rt);
}

LL ask(LL L, LL R, LL l, LL r, LL rt)
{
    if (L <= l&&r <= R) return sum[rt];
    LL mid = (l + r) >> 1;
    LL ans = 0;
    if (L <= mid) ans |= ask(L, R, lson);
    if (R > mid) ans |= ask(L, R, rson);
    return ans;
}

int main()
{
    LL t, n, m;
    cin >> t;
    for (LL i = 1; i <= t; i++){
        scanf("%I64d%I64d", &n, &m);
        build(1, n, 1);
        LL pos = 1;
        LL ans = 0;
        for (LL j = 1; j <= n; j++){
            while (pos <= j&&ask(pos, j, 1, n, 1) >= m) pos++;
            if (pos <= j) ans += (j - pos + 1);
        }
        printf("Case #%I64d: %I64d
",i,ans);
    }
    return 0;
}