IP聚合

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子网掩码： A.B.C.D

IP地址：a.b.c.d

网络地址：(A&a).(B&b).(C&c).(D&d)

问在固定的子网掩码下有多少个网络地址。

#include <cstdio>#include <set>using namespace std;struct rec{    int a, b, c, d;        bool operator <(const rec &s) const    {        if (this->a == s.a && this->b == s.b && this->c == s.c)            return this->d < s.d;        else if (this->a == s.a && this->b == s.b)            return this->c < s.c;        else if (this->a == s.a)            return this->b < s.b;        else            return this->a < s.a;    }};set<rec> s;const int MAXN = 1000;const int MAXM = 50;rec ip[MAXN];rec sm[MAXM];int main(){    int t;    scanf("%d", &t);        for (int i = 1; i <= t; i++) {        printf("Case #%d:\n", i);                int n, m;        scanf("%d%d", &n, &m);                for (int j = 0; j < n; j++) {            scanf("%d.%d.%d.%d", &ip[j].a, &ip[j].b, &ip[j].c, &ip[j].d);        }        for (int j = 0; j < m; j++) {            scanf("%d.%d.%d.%d", &sm[j].a, &sm[j].b, &sm[j].c, &sm[j].d);        }        for (int j = 0; j < m; j++) {            for (int k = 0; k < n; k++) {                rec tmp;                tmp.a = sm[j].a & ip[k].a;                tmp.b = sm[j].b & ip[k].b;                tmp.c = sm[j].c & ip[k].c;                tmp.d = sm[j].d & ip[k].d;                s.insert(tmp);            }            printf("%lu\n", s.size());            s.clear();        }    }        return 0;}

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