UVa 1590

题意

给出几个IP地址
找最小的范围使得输入全部包含在范围之中
输出该范围的最小IP地址和子网掩码

将ip地址和子网掩码看作二进制
子网掩码前32-n位均为1, 后n位为0
对应ip地址范围前32-n位相同, 后n位可以不同

AC代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>

using namespace std;
const int maxn = 1000+10;

int ip[4][maxn];
int s1[4], s2[4];
int mark[] = {255, 254, 252, 248, 240, 224, 192, 128, 0};
//分别是子网掩码中11111111,11111110,11111100,11111000,...,00000000二进制对应值

int main()
{
    int T, i, j, mrk;
    int maxip, minip;
    while( ~scanf("%d",&T))
    {
        memset( ip, 0, sizeof(ip) );
        memset( s1, 0, sizeof(s1) );
        memset( s2, 0, sizeof(s2) );
        for( i = 0; i < T; i++ )
            scanf("%d.%d.%d.%d",&ip[0][i],&ip[1][i],&ip[2][i],&ip[3][i]);
        for( i = 0; i < 4; i++)
        {
            mrk = 0;
            sort(ip[i], ip[i]+T);
            maxip = ip[i][T-1], minip = ip[i][0];
            for(j = 0; j < 8; j++)
            {
                if(maxip % 2 != minip % 2)
                    mrk = j+1;
                maxip /= 2, minip /= 2;
            }
            s2[i] = mark[mrk];
            s1[i] = ip[i][0] & s2[i];
        }
        for( i = 0; i < 4; i++ ){
            if(s2[i] != mark[0]){
                for( i = i + 1; i < 4; i++){
                    s2[i] = 0;
                    s1[i] = 0;
                }
                break;
            }
        }
        printf("%d.%d.%d.%d
",s1[0], s1[1], s1[2], s1[3]);
        printf("%d.%d.%d.%d
",s2[0], s2[1], s2[2], s2[3]);

    }
    return 0;
}