位运算统计组合-优快云博客

本文介绍了一个使用位运算处理数字组合的C++程序。该程序通过输入一系列整数或字符串形式的数字，并利用位运算统计所有可能的组合数量。具体实现包括了位运算的基本操作、动态规划思想的应用以及组合计数的算法。

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#include<iostream>
#include<algorithm>
#include<string.h>
#include<cstdio>
long long a[12], b[12];
int c[1024];
using namespace std;
int main()
{
    b[0] = 1;
    for (int i = 1; i <= 10; i++)
        b[i] = 2 * b[i - 1];
    int n;
    while (scanf("%d",&n)!=EOF)
    {
        memset(c, 0, sizeof(c));
        for (int i = 1; i <= n; i++)
        {
            memset(a, 0, sizeof(a));
            long long s = 0;
            long long m; scanf("%lld", &m);
            while (m > 0)
            {
                int k = m % 10;
                a[k] = 1;
                m = m / 10;
            }
            for (int i = 0; i <= 9; i++)
                s = s + a[i] * b[i];
            c[s]++;
        }
        long long t = 0;
        for (int i = 0; i <= 1023; i++)
        {
            for (int j = i; j <= 1023; j++)
            {
                if (i == j)
                    t = t + c[i] * (c[i] - 1) / 2;
                else
                {
                    if (j&i)
                    {
                        t = t + c[i] * c[j];
                    }
                }
            }
        }
        printf("%lld\n", t);
    }
    return 0;
}

#include<iostream>
#include<algorithm>
#include<string.h>
#include<cstdio>
int a[12], b[12];
int c[1024];
char s[20];
using namespace std;
int main()
{
    b[0] = 1;
    for (int i = 1; i <= 10; i++)
        b[i] = 2 * b[i - 1];
    int n;
    while (scanf("%d", &n) != EOF)
    {
        memset(c, 0, sizeof(c));
        for (int i = 1; i <= n; i++)
        {
            scanf("%s", s);
            int e = 0;
            for (int j = 0; s[j]; j++)
                e = e | b[s[j] - '0'];
            c[e]++;
        }
        long long t = 0;
        for (int i = 0; i <= 1023; i++)
        {
            for (int j = i; j <= 1023; j++)
            {
                if (i == j)
                    t = t + c[i] * (c[i] - 1) / 2;
                else
                {
                    if (j&i)
                    {
                        t = t + c[i] * c[j];
                    }
                }
            }
        }
        printf("%lld\n", t);
    }
    return 0;
}