2014
11-30

# Math teacher’s homework

Mr. Furion is a math teacher. His students are very lazy and they do not like to do their homework. One day, Mr. Furion decides to give them a special problem in order to see whether his students are talents in math or they are just too lazy to do their homework. The problem is:

Given an integer k, n integers m1,m2…mn, and a formula below:

X1 xor X2 xor X3… xor Xn = k

Please figure out that how many integral solutions of the formula can satisfy:

0<=Xi<=mi (i=1…n)

There are at most 100 test cases.

The first line of each test case contains two integers n and k. The second line of each test contains n integers: m1,m2…mn. The meaning of n,k, m1,m2…mn are described above. (1<=n<=50,0<=k,m1,m2…mn<=2^31-1 )

The input is ended by “0 0”.

There are at most 100 test cases.

The first line of each test case contains two integers n and k. The second line of each test contains n integers: m1,m2…mn. The meaning of n,k, m1,m2…mn are described above. (1<=n<=50,0<=k,m1,m2…mn<=2^31-1 )

The input is ended by “0 0”.

11 2047
1024 512 256 128 64 32 16 8 4 2 1
10 2047
1024 512 256 128 64 32 16 8 4 2
0 0

1
0

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#define INF 305
#define inf 0x0f0f0f0f
#define MOD 1000000003LL

using namespace std;
long long F[55][55],Num[55];
int main()
{
int n,k,i,j,t,tot;
long long ans,now;
while (scanf("%d%d",&n,&k),n)
{
for (i=1;i<=n;i++)
scanf("%I64d",&Num[i]);
for (ans=0,t=30;t>=0;t--)
{
memset(F,0,sizeof(F));
for (tot=0,now=1<<t,F[0][0]=i=1;i<=n;i++)
{
if (now&Num[i]) {
for (tot++,j=0;j<tot;j++)
F[i][j]=(F[i-1][j]*(Num[i]-now+1))%MOD;
F[i][1]=(F[i][1]+F[i-1][0])%MOD;
for (j=2;j<=tot;j++)
F[i][j]=(F[i-1][j-1]*(now)+F[i][j])%MOD;
Num[i]-=now;
}
else {
for (j=0;j<=tot;j++)
F[i][j]=(F[i-1][j]*(Num[i]+1))%MOD;
}
}
for (j=1;j<=tot;j++)
if (((tot-j)&1)==((k>>t)&1))
ans=(ans+F[n][j])%MOD;
if (((k>>t)&1)!=(tot&1)) break;
}
if (t==-1) ans=(ans+1)%MOD;
printf("%I64d\n",ans);
}
return 0;
}