HDU-4810-wall Painting(二进制, 组合数)

链接:

https://vjudge.net/problem/HDU-4810

题意:

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.

思路:

将整数转换为二进制存储,每次对二进制的每一位选择,选择奇数个1,xor出来才有值.

每次组合数枚举可选的整数.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;

const int MAXN = 1e3+10;
const int MOD = 1e6+3;
LL a[MAXN];
LL C[MAXN][MAXN];
LL Num[100];
int n;

int main()
{
    C[0][0] = 1;
    C[1][0] = C[1][1] = 1;
    for (int i = 2;i < MAXN;i++)
    {
        C[i][0] = C[i][i] = 1;
        for (int j = 1;j < i;j++)
            C[i][j] = (C[i-1][j]+C[i-1][j-1])%MOD;
    }
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    while (cin >> n)
    {
        memset(Num, 0, sizeof(Num));
        for (int i = 1;i <= n;i++)
        {
            LL v;
            int cnt = 0;
            cin >> v;
            while (v)
            {
                Num[cnt++] += v%2;
                v >>= 1;
            }
        }
        for (int i = 1;i <= n;i++)
        {
            LL res = 0;
            for (int j = 31;j >= 0;j--)
            {
                LL tmp = 0;
                for (int k = 1;k <= i;k += 2)
                    tmp = (tmp + (1LL*C[Num[j]][k]*C[n-Num[j]][i-k])%MOD)%MOD;
                res = (res + (1LL*tmp*(1LL<<j))%MOD)%MOD;
            }
            if (i == n)
                cout << res;
            else
                cout << res << ' ' ;
        }
        cout << endl;
    }

    return 0;
}