You may wonder why this country has such an
interesting tradition? It has a very long story, but I won't tell you
:).
Let us continue, the party princess's knight win the algorithm
contest. When the devil hears about that, she decided to take some
action.
But before that, there is another party arose recently, the
'MengMengDa' party, everyone in this party feel everything is 'MengMengDa' and
acts like a 'MengMengDa' guy.
While they are very pleased about that, it
brings many people in this kingdom troubles. So they decided to stop
them.
Our hero z*p come again, actually he is very good at Algorithm
contest, so he invites the leader of the 'MengMengda' party xiaod*o to compete
in an algorithm contest.
As z*p is both handsome and talkative, he has
many girl friends to deal with, on the contest day, he find he has 3 dating to
complete and have no time to compete, so he let you to solve the problems for
him.
And the easiest problem in this contest is like that:
There
is n number a_1,a_2,...,a_n on the line. You can choose two set
S(a_s1,a_s2,..,a_sk) and T(a_t1,a_t2,...,a_tm). Each element in S should be at
the left of every element in T.(si < tj for all i,j). S and T shouldn't be
empty.
And what we want is the bitwise XOR of each element in S is equal
to the bitwise AND of each element in T.
How many ways are there to
choose such two sets? You should output the result modulo 10^9+7.
number of the test cases.
For each test case, the first line contains a
integers n.
The next line contains n integers a_1,a_2,...,a_n which are
separated by a single space.
n<=10^3, 0 <= a_i <1024,
T<=20.
line.
#include<cstdio>
#include<cstring> typedef __int64 LL;
#define mod 1000000007
const int MAXN = ;
const int MAXA = ;
int dp1[MAXN][MAXA], dp2[MAXN][MAXA], dp3[MAXN][MAXA];
int a[MAXN]; int main()
{
int T, n, i, j, t;
scanf("%d",&T);
while(T--) {
scanf("%d",&n);
for(i = ; i < n; i++)
scanf("%d",&a[i]);
memset(dp1, , sizeof(dp1));
memset(dp2, , sizeof(dp2));
memset(dp3, , sizeof(dp3));
dp1[][a[]] = ;
for(i = ; i < n - ; i++) {
dp1[i][a[i]]++; //单独一个元素构成一个集合
for(j = ; j < MAXA; j++) {
if(dp1[i-][j]) {
dp1[i][j] += dp1[i-][j]; //不添加第i个元素进行异或,继承之前算好的
dp1[i][j] %= mod; t = j ^ a[i]; //添加第i个元素进行异或
dp1[i][t] += dp1[i-][j];
dp1[i][t] %= mod;
}
}
}
dp2[n-][a[n-]] = ;
dp3[n-][a[n-]] = ;
for(i = n-; i > ; i--) {
dp2[i][a[i]]++;
dp3[i][a[i]]++; //单独一个元素构成一个集合
for(j = ; j < MAXA; j++) {
if(dp2[i+][j]) {
dp2[i][j] += dp2[i+][j]; //不添加第i个元素进行按位与
dp2[i][j] %= mod; t = j & a[i]; //添加第i个元素进行按位与
dp2[i][t] += dp2[i+][j];
dp2[i][t] %= mod; dp3[i][t] += dp2[i+][j]; //添加第i个元素进行按位与
dp3[i][t] %= mod;
}
}
}
int ans = ;
for(i = ; i < n - ; i++) {
for(j = ; j < MAXA; j++) {
if(dp1[i][j] && dp3[i+][j]) {
ans += (LL(dp1[i][j]) * dp3[i+][j] % mod);
ans %= mod;
}
}
}
printf("%d\n", ans);
}
return ;
}
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn=1e3+;
const int maxm=*;
const int mod=1e9+;
int n,a[maxn];
int dp[maxn][maxm][],dps[maxn][maxm][];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
int maxi=;
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
memset(dp,,sizeof(dp));
memset(dps,,sizeof(dps));
for(int i=;i<=n;i++)
{
dp[i][a[i]][]=;
dps[i][a[i]][]=;
for(int j=;j<maxm;j++)
if(dp[i-][j][])
{
dp[i][j][]=(dp[i][j][]+dp[i-][j][])%mod;
dp[i][j^a[i]][]=(dp[i][j^a[i]][]+dp[i-][j][])%mod;
dps[i][j^a[i]][]=(dps[i][j^a[i]][]+dp[i-][j][])%mod;
}
}
for(int i=n;i>=;i--)
{
dp[i][a[i]][]=;
for(int j=;j<maxm;j++)
if(dp[i+][j][])
{
dp[i][j][]=(dp[i][j][]+dp[i+][j][])%mod;
dp[i][j&a[i]][]=(dp[i][j&a[i]][]+dp[i+][j][])%mod;
}
}
long long ans=;
for(int i=;i<n;i++)
for(int j=;j<maxm;j++)
if(dps[i][j][])
ans=(ans+(long long)dps[i][j][]*(long long)dp[i+][j][])%mod;
printf("%I64d\n",ans);
}
return ;
}