HDU 4810 Wall Painting

最新推荐文章于 2020-09-15 17:44:35 发布

LZK1997

最新推荐文章于 2020-09-15 17:44:35 发布

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本文介绍了一道关于颜色混合的算法题，通过分析颜色混合规则，采用二进制位操作和组合数学的方法解决该问题。文章详细解释了如何计算不同方案下的颜色总和，并提供了具体的代码实现。

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Wall Painting

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3498 Accepted Submission(s): 1150

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 <= 10 ³).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 10 ⁶ +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

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网上说这是水题，一眼就看出来，我觉得然并卵，作为菜鸟的我，对这种题，也a不掉，一开始想的是每次计算这一层的时候利用上一层的结果，gg，超时，，，，

然后事后看了看他们的博客，才想起来对每一位进行处理，看看这一位中含有多少个1，在看看你需要选出来，用组合数来解

然后枚举1的个数为奇数的情况，注意个数不能大于n和天数，以及选0的个数也不能大于天数。

来个例子把

比如实例

1 2 10 1

化成二进制就是

8 4 2 1

=======================

0 0 0 1

0 0 1 0

1 0 1 0

0 0 0 1

===========================

比如第四位，只含有一个1，如果我们选一种颜料的时候，选择方案在这一位上的表现就是那么出0，要么出1，很明显，出0的时候，对结果值，无贡献，只有出来1的时候，出来一的情况下对结果有贡献，然后出来1有C（1，1）个，乘起来就好啦

My ugly code

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#define ll long long

using namespace std;
const int mod=1e6+3;
int n,a[1010];
int c[1005][1005];
int cnt[35];

int main(){

    /*组合数打表*/
    memset(c,0,sizeof(c));
    c[0][0]=1;
    for(int i=1;i<=1000;i++){
        for(int j=0;j<=i;j++){
            if(j==0 || j==i) c[i][j]=1;
            else c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
        }
    }

    while(~scanf("%d",&n)){
        memset(cnt,0,sizeof(cnt));
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            /*统计输入的每一个数的每一位*/
            for(int j=0;j<=31;j++) {
                if((a[i]>>j)&1) cnt[j]++;
            }
        }
        ll ans[1010];
        /*三层循环，最里面那个是出来1-i个1*/
        for(int i=1;i<=n;i++){
            ans[i]=0;
            for(int j=0;j<=31;j++){
                for(int k=1;k<=cnt[j] && k <= i;k+=2){
                        /*这里为什么这么写，体会下，对每一位处理*/
                    ans[i]=(ans[i]+(ll)(1<<j)%mod*(ll)c[cnt[j]][k]%mod*(ll)c[n-cnt[j]][i-k]%mod)%mod;
                }
                ans[i]%=mod;
            }
        }
        printf("%I64d",ans[1]);
        for(int i=2;i<=n;i++)
            printf(" %I64d",ans[i]);
        printf("\n");
    }


    return 0;
}