2014 ACM/ICPC 아시아 베이징 스팟

2014 ACM/ICPC 아시아 베이징 역(HDU5112-5122)

(이하 제목 설명은 UVA LIVE에서 발췌한 것)

H 문제

제목 설명

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 friend’s 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. Hint: 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. Sample Input 23 2 1 2 3 3 3 1 2 3 Sample Output Case #1: 4 Case #2: 2

제목의 뜻

N을 세어 M보다 크거나 다른 조합이 몇 개 있는지 물어보세요.

해법

각 수와 M이 10^6보다 작거나, 이차 또는 최대 2^20-1이라는 것을 알아차리면 DP를 생각해야 한다.pp[i][j]는 전 i개수에 대해 이차 또는 제이의 조합수를 나타낸다.1차로 스크롤할 수 있는 그룹을 표시합니다.

#include
#include

using namespace std;

int N, M;
typedef long long ll;
const int SIZE = 1 << 20 - 1;
ll dp[2][1 << 20];

int main(){
    int T;
    scanf("%d",&T);
    for(int cs = 1; cs <= T; cs++){
        scanf("%d %d",&N,&M);
        int now = 0, otr;
        memset(dp[now],0,sizeof(dp[now]));
        dp[0][0] = 1;
        for(int i = 1; i <= N; i++){
            int a;
            scanf("%d",&a);
            otr = now;
            now = i & 1;
            memset(dp[now],0,sizeof(dp[now]));
            for(int j = 0; j <= SIZE; j++){
                dp[now][j ^ a] += dp[otr][j];
                dp[now][j] += dp[otr][j];
            }
        }
        ll res = 0;
        for(int i = M; i <= SIZE; i++) res += dp[now][i];
        printf("Case #%d: %lld
",cs,res);
    }
    return 0;
}