hdu 4737 A Bit Fun(TwoPointer)

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题目链接；hdu 4737 A Bit Fun

题目大意：给定一个长度为n的序列,现在问说有多少对i，j满足 f(i,j)<m.

解题思路: Twopointer，将每个数拆分成二进制形式，然后维护连个指针l，r，保证f(l, r) < m，那么当i = l时，对应的j就有r - l + 1种选择方法。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 100005;int c[35], arr[maxn];int add (int x, int v) {    int ret = 0;    for (int i = 0; i <= 30; i++) {        if (x&(1<<i))            c[i] += v;        if (c[i])            ret |= (1<<i);    }    return ret;}ll solve () {    ll ret = 0;    memset(c, 0, sizeof(c));    int l = 0, n, m, x, s = 0;    scanf("%d%d", &n, &m);    for (int i = 0; i < n; i++) {        scanf("%d", &arr[i]);        s = add(arr[i], 1);        while (s >= m)            s = add(arr[l++], -1);        ret += (i - l + 1);    }    return ret;}int main () {    int cas;    scanf("%d", &cas);    for (int kcas = 1; kcas <= cas; kcas++) {        ll ans = solve();        printf("Case #%d: %I64d\n", kcas, ans);    }    return 0;}

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