2013ACM/ICPC亚洲区南京站现场赛 Wall Painting

题意：
给一个长度为$n$的序列，求所有数据$1...n$任意组合异或累加的答案，不考虑顺序
对于$1$，我们只需要求序列的和，对于$2$我们可以找所有两个异或的答案累加，不考虑顺序。

思路：
将每个数处理为2进制串，然后依次枚举二进制串上的每一位，然后利用组合数学求所有异或答案为1的可能，然后乘上$kms(2,j)，j$代表二进制串上的位数

c++不知道为什么$TLE$，g++可过

参考代码：

/*
 * @Author: vain
 * @Date: 2020-08-26 11:58:46
 * @LastEditTime: 2020-09-04 21:37:48
 * @LastEditors: sueRimn
 * @Description: In User Settings Edit
 * @FilePath: \main\demo.cpp
 */
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
//#define ll long long
typedef unsigned long long ull;
const ll N = 2e3 + 20;
const ll maxn = 1e5 + 20;
const ll mod = 1e6 + 3;
ll head[maxn], stac[maxn];
ll a[maxn], b[N], c[maxn], cnt;
//typedef pair p;
//priority_queue, greater
 > m;
ll sum[35][N];
ll max(ll a, ll b) {
      return a > b ? a : b; }
ll min(ll a, ll b) {
      return a < b ? a : b; }
ll gcd(ll a, ll b) {
      return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) {
      return a * b / gcd(a, b); }
void swap(ll &x, ll &y) {
      x ^= y, y ^= x, x ^= y; }
map<string, ll> pll, che, mp;
ll f[N][35], dp[35][2];
int lowbit(int x) {
      return (x) & (-x); }
vector<string> fc;
ull h[maxn], p[maxn];
const ull base = 13331;
char s1[maxn];
ll ksm(ll a, ll b)
{
     
    ll res = 1;
    while (b)
    {
     
        if (b & 1)
            res = res * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return res % mod;
}
ll fac[maxn], inv[maxn];
void init()
{
     
    fac[0] = inv[0] = 1;
    for (int i = 1; i < N; i++)
    {
     
        fac[i] = (fac[i - 1] * i) % mod;
        inv[i] = ksm(fac[i], mod - 2);
    }
}
ll C(ll n, ll m)
{
     
    if (n < 0 || m < 0 || m > n)
        return 0;
    return (fac[n] * inv[m] % mod) * inv[n - m] % mod;
}

void solv(ll x, ll k)
{
     
    for (int i = 31; i >= 0; i--)
        f[k][i] = (x >> i) & 1;
}

ll get(int x, int j)
{
     
    ll ans = 0;
    for (int i = 1; i <= j; i += 2)
    {
     
        ans = (ans + C(dp[x][1], i) * C(dp[x][0], j - i)) % mod;
    }
    return ans;
}
int main()
{
     
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    srand(time(NULL));
    ll n, m, ans, t, x;
    init();
    while (cin >> n)
    {
     
        for (int i = 0; i < 33; i++)
            dp[i][0] = dp[i][1] = 0;
        for (int i = 0; i < n; i++)
        {
     
            cin >> x, solv(x, i);
        }
        for (int i = 0; i < 32; i++)
        {
     
            for (int j = 0; j < n; j++)
            {
     
                dp[i][f[j][i]]++;
            }
        }
        for (int i = 0; i < 32; i++)
        {
     
            for (int j = 1; j <= n; j++)
            {
     
                sum[i][j] = get(i, j);
            }
        }
        for (int i = 1; i <= n; i++)
        {
     
            ll ans = 0;
            for (int j = 0; j < 32; j++)
            {
     
                ans = (ans + ksm(2, j) * sum[j][i]) % mod;
            }
            if (i != n)
                cout << ans << ' ';
            else
                cout << ans << endl;
        }
    }
}