HDU 4737 A Bit Fun

题目链接

由于位运算|在区间上是单调的，所以只需要算出每个从i开始的能到哪个j结束，使得(i,j)这个区间里面所有子区间都是满足条件的，即可。

 1 #include <stdio.h>
 2 #include <string.h>
 3 typedef long long ll;
 4 const int maxn = 1e5 + 5;
 5 int n, m, a[maxn], b[31];
 6 inline int modify(int x, int v) {
 7     int ret = 0;
 8     for (int i = 0; i <= 30; ++i, x >>= 1) {
 9         if (x & 1)
10             b[i] += v;
11         if (b[i])
12             ret |= 1 << i;
13     }
14     return ret;
15 }
16 int main() {
17     int T;
18     scanf("%d", &T);
19     for (int Tc = 1; Tc <= T; ++Tc) {
20         scanf("%d%d", &n, &m);
21         for (int i = 0; i < n; ++i)
22             scanf("%d", &a[i]);
23         memset(b, 0, sizeof(b));
24         int j = 0, tmp = 0;
25         ll ans = 0;
26         for (int i = 0; i < n; ++i) {
27             if (i)
28                 tmp = modify(a[i - 1], -1);
29             while (j <= i || (j < n && (tmp | a[j]) < m))
30                 tmp = modify(a[j++], 1);
31             if (tmp < m)
32                 ans += j - i;
33         }
34         printf("Case #%d: %I64d\n", Tc, ans);
35     }
36     return 0;
37 }