HDU 47,372 points or violence

http://acm.hdu.edu.cn/showproblem.php?pid=4737

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 Input

23 61 3 52 45 4

Sample Output

Case #1:4Case #2:0

<pre name= "code" class= "CPP" &GT;/**HDU 47,372 points or the topic of violence: Given a column number, find out how many sub-range satisfies all the numbers in the interval or the value is less than M solution thinking: My method is to open a number num[n][35],num[ I][J] Indicates how many of the first I number of J bits on the number of 1, and then starting from 1 to enumerate the beginning of the interval, the binary search does not meet the first point less than M can be, the complexity O (NLOGN) teammates are direct violence O (n*n), do not know why faster than mine ==*/# Include <stdio.h> #include <string.h> #include <iostream> #include <algorithm>using namespace    std;const int Maxn=100005;int n,m;int num[maxn][35],a[maxn],bit[35];int judge (int n,int x) {int sum=0;    for (int i=0;i<33;i++) {if (num[n][i]-num[x-1][i]>0) sum+= (1<<i); } return sum;}    int erfen (int x) {int l=x,r=n,mid;        while (l<=r) {mid= (l+r)/2;        if (judge (mid,x) >=m) r=mid-1;    else l=mid+1; } return R;}    int main () {int t,tt=0;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n,&m);        memset (num,0,sizeof (num));            for (int i=1;i<=n;i++) {scanf ("%d", &a[i]); int x=a[i],len=0;            for (int j=0;j<33;j++)////This sentence is very important, do not use Num[i][len]=num[i-1][len]bit[len under the following]; Because Len of I does not necessarily have i-1 big num[i            ][J]=NUM[I-1][J];                while (x) {bit[len]=x&1;                Num[i][len]+=bit[len];                len++;            x>>=1;                }}/** for (int i=1;i<=n;i++) {for (int j=0;j<30;j++) {            printf ("%d", num[i][j]);        } printf ("\ n");        }*/int ans=0;            for (int i=1;i<=n;i++) {int Y=erfen (i);            ans+= (y-i+1);        printf ("%d%d\n", i,y);    } printf ("Case #%d:%d\n", ++tt,ans); } return 0;} /**998 81 2 3 4 5 6 7 88 71 2 3 4 5 6 7 84 41 2 3 410 232 3 5 7 8 1 2 4 15 57*//***//attached violence code # include <iostream> #inclu    De <cstdio>using namespace Std;const int maxn = 1e5+10;int A[maxn];int main () {int t,cas=1,n,m;    scanf ("%d", &t); while (t--) {scanf ("%d%d ", &n,&m);        for (int i=0;i<n;i++) scanf ("%d", a+i);        int ans = 0;            for (int i=0;i<n;i++) {int tmp = A[i];            if (tmp<m) ans++;                for (int j=i+1;j<n;j++) {tmp|=a[j];                if (tmp<m) ans++;            else break;    }} printf ("Case #%d:%d\n", Cas++,ans); } return 0;} */

HDU 47,372 points or violence