2
5 2
192.168.1.0
192.168.1.101
192.168.2.5
192.168.2.7
202.14.27.235
255.255.255.0
255.255.0.0
4 2
127.127.0.1
10.134.52.0
127.0.10.1
10.134.0.2
235.235.0.0
1.57.16.0

题目意思很清楚，只要使用按位与的的方法，计算出相应结果就可以了。。

水题一道，可以练练常用的数据结构...

#include <cstdio>
#include <vector>
#include <set>
#include <iostream>
#include <cstring>
using namespace std;
const int MAXN = 1005;
struct IntNode{int a,b,c,d;};

IntNode ym[MAXN];      //掩码
IntNode ip[MAXN];      //ip地址

set<long long> res;
int T,N,M;
int cas=1;
long long calculate(int i,int j)
{
    long long tmpa,tmpb,tmpc,tmpd,sum;
    tmpa = ym[i].a & ip[j].a;
    tmpb = ym[i].b & ip[j].b;
    tmpc = ym[i].c & ip[j].c;
    tmpd = ym[i].d & ip[j].d;
    sum = tmpd + tmpc * 1000 + tmpb * 1000000 + tmpa * 1000000000;
    return sum;
}
int main()
{
    cin>>T;
    while(cin>>N>>M)
    {
        memset(ym,sizeof(ym));
        memset(ip,sizeof(ip));
        for(int i = 1;i <= N; i++)
            scanf("%d.%d.%d.%d",&ip[i].a,&ip[i].b,&ip[i].c,&ip[i].d);
        for(int i = 1;i <= M; i++)
            scanf("%d.%d.%d.%d",&ym[i].a,&ym[i].b,&ym[i].c,&ym[i].d);

        printf("Case #%d:\n",cas++);
        for (int i = 1; i <= M; i++)
        {
            res.clear();
            for(int j = 1; j <= N; j++)
            {
                long long num = calculate(i,j);
                if (!res.count(num))
                    res.insert(num);
            }
            printf("%d\n",res.size());
        }
    }
    return 0;
}