Happy Matt Friends
Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 150 Accepted Submission(s): 56
Problem Description
Matt has N friends. They are playing a game together.
Each of Matt’s friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends’magic numbers is no less than M , Matt wins.
Matt wants to know the number of ways to win.
Input
The first line contains only one integer T , which indicates the number of test cases.
For each test case, the first line contains two integers N, M (1 ≤ N ≤ 40, 0 ≤ M ≤ 106).
In the second line, there are N integers ki (0 ≤ ki ≤ 106), indicating the i-th friend’s magic number.
Output
For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y indicates the number of ways where Matt can win.
Sample Input
2 3 2 1 2 3 3 3 1 2 3
Sample Output
Case #1: 4 Case #2: 2HintIn the first sample, Matt can win by selecting: friend with number 1 and friend with number 2. The xor sum is 3. friend with number 1 and friend with number 3. The xor sum is 2. friend with number 2. The xor sum is 2. friend with number 3. The xor sum is 3. Hence, the answer is 4.
Source
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题解是高斯消元（真心不太会）
用了背包过的，再加个滚动数组，一开始纠结在前i个数异或最大，然后怎么都A不了，后来干脆全部把上限改成1 << 20，然后过了
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
__int64 dp[2][1 << 22];
int num[44];
int main()
{
int t, n, m, icase = 1;
__int64 ans;
scanf("%d", &t);
while (t--)
{
memset (dp, 0, sizeof(dp));
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i)
{
scanf("%d", &num[i]);
}
dp[0][0] = 1;
for (int i = 1; i <= n; ++i)
{
for (int j = 0; j <= (1 << 20); ++j)
{
dp[i % 2][j] = dp[1 - (i % 2)][j] + dp[1 - (i % 2)][j ^ num[i]];
}
}
ans = 0;
for (int i = m; i <= (1 << 20); ++i)
{
ans += dp[n % 2][i];
}
printf("Case #%d: %I64d\n", icase++, ans);
}
return 0;
}