HDU 4810 Wall Painting（组合数学 + 位运算）——2013ACM/ICPC

发布时间：2020-12-13 21:16:05 所属栏目：PHP教程来源：网络整理

导读：传送门 Wall Painting Time Limit: 10000/5000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2627Accepted Submission(s): 839 Problem Description Ms.Fang loves painting very much. She paints GFW(Great Funny Wal

传送门

Wall Painting

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2627 Accepted Submission(s): 839

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

题目大意：

给定 n 个数，求任意 ∑ni=1C(n,i) 个数的异或和。

解题思路：

首先将每个数的2进制求出来，在每位2进制数中求出 1 的个数，拿样例来讲吧。

1  =  0   0   0   1
2  =  0   0   1   0
10=  1   0   1   0
1  =  0   0   0   1

每次选奇数个 1 和剩余的选的个数⑴ 个 0，比如说求任意两个数的异或和，那末我们选 1 个 1 ，选 2?1 个 0，我们肯定选

奇数个 1，由于偶数个 1 异或还是 0 没成心义，然后就是组合数了，从当前的 1 的总个数里面选择奇数个 1,在乘以相应的 2x

次幂，(c[cnt[j]][k]?c[(n?cnt[j])][i?k])?(2j)其中 cnt[j] 表示的是第 j 位 1 的总个数， k 表示的是要选择的 1 的个数。i 表示的

是任意 i 个数异或。。。大体上看看代码就好懂了。

My Code：

/**
2016 - 09 - 20 晚上
Author: ITAK
Motto:

本日的我要超出昨日的我，明日的我要胜过本日的我，
以创作出更好的代码为目标，不断地超出自己。
**/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 1e9+5;
const int MAXN = 1e3+5;
const int MOD = 1e6+3;
const double eps = 1e⑺;
const double PI = acos(-1);
using namespace std;

int Scan_Int()///输入外挂
{
    int res = 0,ch,flag = 0;
    if((ch=getchar()) == '-')
        flag = 1;
    else if(ch >= '0' && ch<='9')
        res = ch-'0';
    while((ch=getchar())>='0' && ch<='9')
        res = res*10+ch-'0';
    return flag?-res:res;
}
LL Scan_LL()///输入外挂
{
    LL res=0,flag=0;
    if((ch=getchar())=='-')
        flag=1;
    else if(ch>='0'&&ch<='9')
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}

void Out(int a)///输出外挂
{
    if(a>9)
        Out(a/10);
    putchar(a%10+'0');
}
LL c[MAXN][MAXN];
void Init()
{
    c[0][0] = 1;
    for(int i=1; i<MAXN; i++)
        c[i][0] = 1;
    for(int i=1; i<MAXN; i++)
        for(int j=1; j<MAXN; j++)
            c[i][j] = (c[i-1][j-1] + c[i-1][j]) % MOD;
}
LL a[MAXN];
int cnt[70];
int main()
{
    Init();
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<n; i++)
            scanf("%I64d",&a[i]);
        memset(cnt,0,sizeof(cnt));
        for(int i=0; i<n; i++)
        {
            LL tp = a[i];
            for(int j=0; j<63; j++)
            {
                if(tp & 1)
                    cnt[j]++;
                tp>>=1;
            }
        }
        for(int i=1; i<=n; i++)
        {
            LL sum = 0;
            for(int j=0; j<63; j++)
            {
                if(cnt[j])
                {
                    for(int k=1; k<=i; k+=2)
                    {
                        if(cnt[j]>=k && (n-cnt[j]>=i-k))
                        {
                            sum += ((c[cnt[j]][k]*c[(n-cnt[j])][i-k]%MOD)*(1LL<<j)%MOD);
                            sum %= MOD;
                        }
                    }
                }
            }
            if(i != n)
                cout<<sum<<" ";
            else
                cout<<sum<<endl;
        }
    }
    return 0;
}

（编辑：李大同）

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