hdu 4810 Wall Painting（二进制+组合数学）

Wall Painting

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 410    Accepted Submission(s): 120

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

题意：求n个数里面，取i个数异或的所有组合的和，i取1~n

题解：将n个数拆成30位2进制，由于每个二进制位异或后相加和原来的数异或相加是一样的，所以只需要对每一位累加计算，用组合数学取数就行了，奇数个异或得1，偶数个异或得0，再乘以自己的二进制位值，复杂度O（30*n*n）

#include<stdio.h>#include<string.h>const int mod = 1000003;int c[1005][1005],po[33],a[33];void init(){    int i,j;    memset(c,0,sizeof(c));    c[0][0]=c[1][0]=c[1][1]=1;    for(i=2; i<=1000; i++)    {        c[i][0]=c[i][i]=1;        for(j=1; j<i; j++)        {            c[i][j]=c[i-1][j-1]+c[i-1][j];            if(c[i][j]>=mod) c[i][j]%=mod;        }    }    po[0]=1;    for(i=1; i<30; i++) po[i]=(2*po[i-1])%mod;}void myadd(int x){    int i=0;    while(x)    {        if(x&1) a[i]++;        x>>=1;        i++;    }}int main(){    int res,n,i,j,k,x;    long long temp;    init();    while(scanf("%d",&n)>0)    {        memset(a,0,sizeof(a));        for(i=0; i<n; i++)        {            scanf("%d",&x);            myadd(x);        }        for(j=1; j<=n; j++)        {            for(res=i=0; i<30; i++)            {                for(k=1;k<=j;k+=2)                {                    temp=(long long)po[i]*c[a[i]][k]*c[n-a[i]][j-k];                    if(temp>=mod) temp%=mod;                    res=res+temp;                    if(res>=mod) res%=mod;                }            }            printf("%d",res);            if(j!=n) printf(" ");            else printf("\n");        }    }    return 0;}