POJ 2799 IP Networks

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;
}