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：

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#include 
  
   
#include 
   
     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<
    
     > 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; }
    
   
  

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