Hdu 4737 A Bit Fun Ruler

A Bit Fun

Time limit:5000/2500 MS (java/others) Memory limit:32768/32768 K (java/others)

Problem descriptionthere is n numbers in a array, as A0, A1 ..., an-1, and another number M. We define a function f (i, j) = ai|ai+1|ai+2| ... | Aj. Where "|" is the bit-or operation. (I <= J)
The problem is really simple:please count the number of the different pairs of (I, J) where F (i, J) < M.

Inputthe first line has a number T (T <=), indicating the number of test cases.
For each test case, first line contains, numbers n and M. (1 <= n <= 100000, 1 <= m <=) then n numbers c Ome in the second line which is the array a, where 1 <= ai <= 230.

Outputfor every case, you should output ' case #t: ' At first, without quotes. The TIs the case number starting from 1.
Then follows the answer.

Sample Input23 61 3 52 45 4

Sample outputcase #1:4Case #2:0

Source

ACM/ICPC Asia Regional Chengdu Online

#include <bits/stdc++.h>using namespacestd;#definell __int64#defineESP 0.00000000001Const intn=1e5+Ten, m=1e6+Ten, inf=1e9+Ten, mod=1000000007;intA[n];intflag[ +];voidinit () {memset (flag,0,sizeof(flag));}voidUpdateintXinthh) {    intsum=0;  while(x) {flag[sum++]+=x%2*hh; X>>=1; }}intGetnum () {intans=0;  for(intI=0; i<= *; i++)    {        if(Flag[i]) ans+=1<<i; }    returnans;}intMain () {intx,y,z,i,t; intt,cas=1; scanf ("%d",&T);  while(t--) {init (); scanf ("%d%d",&x,&y);  for(i=1; i<=x;i++) scanf ("%d",&A[i]); intst=1; inten=1; intCnt=1; ll ans=0; intqu=0;  while(1)        {             while(Getnum () <y&&en<=x) Update (A[en++],1); if(Getnum () <y) {ans+ = (ll) (en-st) * (en-st+1)/2;  Break; } ans+=max (0, en-st-1); Update (A[st++],-1); } printf ("Case #%d:%i64d\n", cas++, ans); }    return 0;}

Hdu 4737 A Bit Fun Ruler