HDU Problem - 4810 Wall Painting【组合数学】

 

Wall Painting

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2706    Accepted Submission(s): 865

Problem Description

Ms.Fang loves painting very much. She paints GFW(Great Funny Wall) every day. Every day before painting, she produces a wonderful color of pigments by mixing water and some bags of pigments. On the K-th day, she will select K specific bags of pigments and mix them to get a color of pigments which she will use that day. When she mixes a bag of pigments with color A and a bag of pigments with color B, she will get pigments with color A xor B.
When she mixes two bags of pigments with the same color, she will get color zero for some strange reasons. Now, her husband Mr.Fang has no idea about which K bags of pigments Ms.Fang will select on the K-th day. He wonders the sum of the colors Ms.Fang will get with different plans.

For example, assume n = 3, K = 2 and three bags of pigments with color 2, 1, 2. She can get color 3, 3, 0 with 3 different plans. In this instance, the answer Mr.Fang wants to get on the second day is 3 + 3 + 0 = 6.
Mr.Fang is so busy that he doesn’t want to spend too much time on it. Can you help him?
You should tell Mr.Fang the answer from the first day to the n-th day.

 

 

Input

There are several test cases, please process till EOF.
For each test case, the first line contains a single integer N(1 <= N <= 10³).The second line contains N integers. The i-th integer represents the color of the pigments in the i-th bag.

 

 

Output

For each test case, output N integers in a line representing the answers(mod 10⁶ +3) from the first day to the n-th day.

 

 

Sample Input

4 1 2 10 1

 

 

Sample Output

14 36 30 8

 

 

Source

2013ACM/ICPC亚洲区南京站现场赛——题目重现

 

 

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题意：

有n个数、n天。第i天从n个数中选出i个数进行按位与运算，求出每天这些数按位与的和。

思路：

任意多个数进行按位与，只要某一位有奇数个1，就可以一把这个数保存下来。因为是在所有的数中任意取，所以只要统计这些数的二进制位，求出每个值的贡献。然后用组合。快速幂+费马小定理会超时，只能用杨辉三角。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const long long MOD = 1e6 + 3;
const int MAXN = 1000 + 10;
LL fac[MAXN];
LL C[MAXN][MAXN];
int bit[32];
void get_c() {
    for(int i = 0; i <= 1000; i++)   C[i][0] = 1;
    for(int i = 1; i <= 1000; i++) {
        for(int j = 1; j <= i; j++) {
            C[i][j] = (C[i-1][j] + C[i-1][j-1]) % MOD;
        }
    }
}
LL myc(int x, int y) {return C[x][y];}
int main() {
    int n, t;
    get_c();
    while (scanf("%d", &n) != EOF) {
        memset(bit, 0, sizeof(bit));
        for (int i = 0; i < n; i++) {
            scanf("%d", &t);
            for (int j = 0; j <= 30; j++) {
                if (t & (1<<j)) bit[j]++;
            }
        }
        for (int k = 1; k <= n; k++) {
            LL ans = 0;
            for (int i = 0; i <= 30; i++) {
                if (!bit[i]) continue;
                for (int j = 1; j <= k && j <= bit[i]; j += 2) {
                    ans += myc(bit[i], j)*myc(n - bit[i], k - j)%MOD* (1<<i)%MOD;
                    ans %= MOD;
                }
            }
            if (k != n) printf("%lld ", ans%MOD);
            else printf("%lld\n", ans%MOD);
        }
    }
    return 0;
}