2014ACM/ICPC亚洲区北京站

2014ACM/ICPC亚洲区北京站(HDU5112-5122)

(以下题目描述摘自UVA LIVE)

H题

题目描述

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

题意

给N个数，问有多少种组合，使他们的异或和大于M。

解法

注意到每个数和M小于10^6，异或和最大为2^20-1，就应该想到DP！
dp[i][j]表示对于前i个数，表示异或和为j的组合数。显示第一维可以滚动数组。

#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\n",cs,res);
    }
    return 0;
}