Topic links: HDU 5119
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 \(N, M (1 \le N \le 40, 0 \le M \le 10^6)\).
In the second line, there are \(N\) integers \(k_i (0 ≤ k_i ≤ 10^6)\), 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 3Sample Output
Case #1: 4
Case #2: 2Hint
In 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
Solution
The meaning of problems
Given \ (n-\) number \ (K [I] \) , which took some number greater than or equal to that exclusive \ (m \) , find several emulated.
Thinking
DP rolling backpack array
Set \ (dp [i] [j ] \) represents the former \ (I \) number and for the exclusive-OR \ (J \) all emulated. State transition equation is \ (DP [I] [J] DP = [I -. 1] [J] + DP [I -. 1] [J \ XOR \ K [I]] \) .
Since only the current state and the previous state, and therefore can be used to optimize the rolling array.
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1 << 20;
ll dp[10][maxn];
ll k[50];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int T;
cin >> T;
for(int _ = 1; _ <= T; ++_) {
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; ++i) {
cin >> k[i];
}
for(int i = 1; i <= n; ++i) {
for(int j = 0; j < maxn; ++j) {
dp[i & 1][j] = dp[(i - 1) & 1][j] + dp[(i - 1) & 1][j ^ k[i]];
}
}
ll ans = 0;
for(int i = m; i < maxn; ++i) {
ans += dp[n & 1][i];
}
cout << "Case #" << _ << ": " << ans << endl;
}
return 0;
}