题意:给出n个数,求取k个数的异或和是多少,k从1到n,分别输出n个答案;

思路:由于这个是异或和,那么我们可以从这里下手,我们把n个数全都看成二进制的形式,那么当前某一位会被计算进总和,当且仅当在所取的k个数中,这一位为1(二进制形式下)的数量是奇数

那么最后答案就是求一个组合数,乘上当前这一位的权值(即2的幂次),之后相加就行了

#include <bits/stdc++.h>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,(rt<<1)+1
#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
const int mod = 1e6 + 3;
int n;
int a[maxn];
ll C[1005][1005];
void init(int N) {//预处理组合数
    for (int i = 0; i < N; i ++) {
        C[i][0] = C[i][i] = 1;
        for (int j = 1; j < i; j ++) {
            C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
            C[i][j] %= mod;
        }
    }
}

int one[1004], zero[1004]; //分别表示第i位有几个
ll ans[maxn];
int main() {
    init(1002);
    while (~scanf("%d", &n)) {
        CLR(one, 0);
        CLR(zero, 0);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            ans[i] = 0;
        }
        for (int i = 0; i < 32; i++) {//预处理第i位1的个数和0的个数
            for (int k = 1; k <= n; k++) {
                int x =  (a[k] >> i) & 1;
                if (x == 1) one[i]++;
                else zero[i]++;
            }
        }
        for (int j = 1; j <= n; j++) {//选j个
            for (int i = 0; i < 32; i++) {//int数据范围下有32位
                int x = one[i];
                int y = zero[i];
                for (int k = 1; k <= j && x >= k; k += 2) {//因为是求异或,所以有贡献的话就要奇数
                    ans[j] = (ans[j] + C[x][k] * C[y][j - k] % mod * (1LL << i)) % mod;
                    //求组合数在乘上一个权值
                }
            }
        }
        for (int i = 1; i <= n; i++) {
            printf("%lld%c", ans[i], i == n ? '\n' : ' ');
        }
    }
    return 0;
}