HDU 4810 Wall Painting （2进制+组合数）

Wall Painting

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3244    Accepted Submission(s): 1068

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

41 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：

给出n种颜色，求1到n种颜色混合产生的颜色种类和。

颜色混合就是对应数字异或；

第k个答案就是在n种颜色中随机挑选k个进行混合的全部结果之和。

思路：

二进制拆位，一位一位计算， 把n个数拆成32位，记录每一位有几个1， 从n个数选k个数的抑或和， k个数每一位有用的其实就是有奇数个1，这样这一位抑或和才有用，然后这一位的答案是 2^pos，枚举k个数内所有奇数个数，计算这一位的总贡献， 总的答案就是每一位的贡献相加。

#include<bits/stdc++.h>using namespace std;typedef long long ll;const int mod = 1e6+3;const int maxn = 1e3+5;ll a[maxn], c[maxn][maxn], n;ll fac[maxn] = {1};ll rec[maxn], b[maxn];ll ans[maxn];void init(){    c[0][0] = 1;    for(int i = 1; i < maxn; i++)        for(int j = 0; j <= i; j++)        {            if(!j || i == j)                c[i][j] = 1;            else                c[i][j] = (c[i-1][j]+c[i-1][j-1])%mod;        }    for(int i = 1; i < maxn; i++)        fac[i] = fac[i-1]*2%mod;}void solve(int k){    memset(rec, 0, sizeof(rec));    int num0 = 0, num1 = 0;    for(int i = 1; i <= n; i++)        b[i] = a[i];    for(int i = 0; i < 33; i++)    {        num1 = num0 = 0;        for(int j = 1; j <= n; j++)        {            int idx = b[j]%2;            if(idx) num1++;            else num0++;            b[j] /= 2;        }        for(int j = 1; j <= k; j += 2)            rec[i] = (rec[i]+c[num1][j]*c[num0][k-j]%mod)%mod;    }    ll res = 0;    for(int i = 0; i < 33; i++)        res = (res+rec[i]*fac[i]%mod)%mod;    ans[k] = res;}int main(void){    init();    while(cin >> n)    {        for(int i = 1; i <= n; i++)            scanf("%lld", &a[i]);        for(int i = 1; i <= n; i++)            solve(i);        for(int i = 1; i <= n; i++)            printf("%lld%c", ans[i], i==n ? '\n' : ' ');    }    return 0;}