首页 > ACM题库 > HDU-杭电 > HDU 4810-Wall Painting-位运算-[解题报告]HOJ
2015
09-18

HDU 4810-Wall Painting-位运算-[解题报告]HOJ

Wall Painting

问题描述 :

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 withCirno’s Present 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.

输入:

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.

输出:

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.

样例输入:

4
1 2 10 1

样例输出:

14 36 30 8

题目:

http://acm.hdu.edu.cn/showproblem.php?pid=4810

题意:

给出n种颜色,求1到n种颜色混合产生的颜色种类和。

颜色混合就是对应数字异或;

第k个答案就是在n种颜色中随机挑选k个进行混合的全部结果之和。

思路:

位运算,按位异或之和等于各数字异或和;

数据小,直接打表组合数,然后按位求得所有可能数,详见code;

代码:

#define MOD 1000003
#define N 1123

int n,m;
int flag,res;
long long sum,ans,t;
long long a[N],b[N];
long long f[N][N],c[N][N];

void init()
{
    memset(f,0,sizeof(f));
    for(int i=0;i<N;i++)
        f[i][0]=f[i][i]=1;
    for(int i=1;i<N;i++)
        for(int j=1;j<i;j++)
            f[i][j] = (f[i-1][j]+f[i-1][j-1])%MOD;
}
int main()
{
    int i,j,k;
    init();
    while(scanf("%d",&n)!=EOF&&n)
    {
        memset(b,0,sizeof(b));
        res=0;
        for(i=0;i<n;i++)
        {
            scanf("%lld",&t);
            j=0;
            while(t)
            {
                if(t&1)
                    b[j]++;
                j++;
                t>>=1;
            }
        }
        for(k=1;k<=n;k++)
        {
            sum = 0;
            t = 1;
            for(i=0;i<32;i++)
            {
                ans=0;
                for(j=1;j<=k;j+=2)
                    ans=(ans+(f[b[i]][j]*f[n-b[i]][k-j])%MOD)%MOD;
                sum=(sum+ans*t%MOD)%MOD;
                t*=2;
            }
            if(k!=1)printf(" ");
            printf("%lld",sum);
        }
        printf("\n");
    }
    return 0;
}

参考:http://blog.csdn.net/kopyh/article/details/47174983