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\n",i,ans);

    }

    return 0;

}