network address是前(32-n)随意 后n位全零

network mask是前(32-n)全一 后n位全零

本题主要利用位移操作，1ULL表示无符号长整型的常数1，这样写可防止不必要的溢出，取反后可以作为mask的枚举然后拿mask和mins或者maxs并一下就得到address了。

代码：

#include <cstdio>

#include <iostream>

using namespace std;

const int maxn = 70;

int main()

{

    int n;

    while(scanf("%d", &n) == 1)

    {

        unsigned int maxs = 0;

        unsigned int mins = 0-1;

        //printf("%u\n",mins);

        int a, b, c, d;

        unsigned int e;

        for(int i = 0; i < n; i++)

        {

            scanf("%d.%d.%d.%d", &a, &b, &c, &d);

            e = ((unsigned int)a << 24) + (b << 16) + (c << 8) + d;

            //printf("%u\n",e);

            if(e < mins)

                mins = e;

            if(e > maxs)

                maxs = e;

        }

        unsigned int mask;

            for (int i = 0; i <= 32; i++)

            {

                mask = ~((1ULL<<i)-1U);//注意这里ULL，是为了防止数据溢出

                //printf("%u\n",mask);

                if ((mins & mask) == (maxs & mask))

                    break;

            }

        unsigned int ans = mins & mask;

        printf("%u.%u.%u.%u\n", ans >> 24, (ans << 8) >> 24, (ans << 16) >> 24, (ans << 24) >> 24);

        printf("%u.%u.%u.%u\n", mask >> 24, (mask << 8) >> 24, (mask << 16) >> 24, (mask << 24) >> 24);

    }

    return 0;

}