HDU - 4810 Wall Painting（组合数学+枚举）

最新推荐文章于 2019-05-14 08:17:17 发布

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最新推荐文章于 2019-05-14 08:17:17 发布

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分类专栏： ACM-数学文章标签：组合数学

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Ms. Fang每天从不同的颜料袋中选择特定数量的颜料，通过混合获得独特的颜色来绘画她的伟大趣味墙（GFW）。她丈夫Mr. Fang想知道在第K天，当Ms. Fang选择K个特定的颜料袋进行混合时，所有可能颜色组合的总和。此问题涉及组合数学和二进制位操作，通过对每种颜料颜色的二进制表示进行分析，可以高效地计算出答案。

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Wall Painting

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3978 Accepted Submission(s): 1317

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

这道题并不好想啊，训练的时候老想异或的性质了，奇数个1异或为1，偶数个1异或为0，怎么也没把时间复杂度降下来，题解利用到了组合数，把每个数化为2进制，统计每一位二进制所有数1的个数，一位一位的计算和，比如样例：

0 0 0 1

1 0 1 0

0 0 1 0

在第2天的时候，答案为 C（1,1） * C（3,1）*（1<<3） + 0 + C（2,1）*C（2,1） * （1<<1）+ C（2,1）*C（2,1） * （1<<0）

利用组合数巧妙的把复杂度降为30n*n了

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1010;
const int MOD = 1e6 + 3;
int num[35];
ll C[MAXN][MAXN];
//****************预处理出小范围的组合数******************//
void init()
{
    int i, j;
    memset(C, 0, sizeof(C));
    for(i = 0; i < MAXN - 5; i++)
        C[i][i] = C[i][0] = 1;
    for(i = 1; i < MAXN - 5; i++)
    for(j = 1; j < i; j++)
        C[i][j] = (C[i-1][j-1] + C[i-1][j]) % MOD;
}
//****************预处理出小范围的组合数******************//
int main(void)
{
    int n,cnt,t;
    ll tt;
    init();
    while(scanf("%d",&n) != EOF) {
        memset(num,0,sizeof(num));
        for(int i = 1; i <= n; i++) {
            scanf("%d",&t);
            cnt = 0;
            while(t) {
                if(t & 1) num[cnt]++;
                t >>= 1;
                cnt++;
            }
        }
        for(int i = 1; i <= n; i++) {
            ll ans = 0;
            for(int j = 0; j <= 30; j++) {
                for(int k = 1; k <= i; k += 2) {
                    tt = (1ll << j) % MOD * C[num[j]][k] % MOD * C[n - num[j]][i - k] % MOD;
                    ans = (ans + tt) % MOD;
                }
            }
            if(i == n) printf("%I64d\n",ans);
            else printf("%I64d ",ans);
        }
    }
    return 0;
}