UVA 1590 IP地址

这道题先求出子网掩码，然后将最小的IP地址与子网掩码按位与即可得到网络地址。注意以下问题：
1.求子网掩码，即所有32位地址最左一个不同出现的地方后全为0，前面全为1。
2.由于IP地址可能有重复，可以用set来保存所有IP地址，从而去除重复。
3.只需要将最小的IP地址与子网掩码按位与即可得到网络地址，set里面是按字典序排的，从而只需要取set第一个IP地址即可。

代码如下：

#include<cstdio>#include<iostream>#include<algorithm>#include<set>#include<cstring>#include<string>using namespace std;int a[4];int b[32];int c[4];int d[32];int ans;int n;set<string> s;void init(){    memset(b, 0, sizeof(b));    for (int i = 0; i < 4; i++){        a[i] = 255;    }    s.clear();    ans = 33;}bool getInput(int a[]){    string str;    cin >> str;    if (s.count(str))        return false;    s.insert(str);    int i = 0;    int cnt = 0;    while (str[i] != '\0'){        int t = 0;        while (str[i] != '.' && str[i]!='\0'){            t = t * 10 + (str[i] - '0');            i++;        }        a[cnt++] = t;        if (str[i] == '\0')            break;        i++;    }    return true;}void convert(){    int t[4];    memcpy(t, c, sizeof(c));    for (int i = 0; i < 4; i++){        for (int j = 0; j < 8; j++){            d[i * 8 +7-j] = t[i]&1;            t[i] = t[i] >> 1;        }    }    if (ans == 33){        memcpy(b, d, sizeof(d));        ans = 32;        return;    }    int i;    for (i = 0; i < ans; i++){        if (d[i] != b[i])            break;    }    if (i < ans)        ans = i;}int myprint(int x){    int t = 0;    for (int i = 0; i < 8; i++)        t = t * 2 + b[i+x];    return t;}void print(){//最终b存放子网掩码地址，a存放网络地址    memset(b, 0, sizeof(b));    for (int i = 0; i < ans; i++)        b[i] = 1;    for (int i = 0; i < 4; i++){        c[i] = myprint(i * 8);    }    string str = *(s.begin());    int i = 0;    int cnt = 0;    while (str[i] != '\0'){        int t = 0;        while (str[i] != '.' && str[i] != '\0'){            t = t * 10 + (str[i] - '0');            i++;        }        a[cnt++] = t;        if (str[i] == '\0')            break;        i++;    }    for (int i = 0; i < 4; i++)        a[i] = a[i] & c[i];    for (int i = 0; i < 4; i++){        if (i)            cout << ".";        cout << a[i];    }    cout << endl;    for (int i = 0; i < 4; i++){        if (i)            cout << ".";        cout << c[i];    }    cout << endl;}int main(){    int m;    while (cin >> m){        init();        n = 0;        for (int i = 0; i < m; i++){            if (!getInput(c))                continue;            convert();        }        print();    }    return 0;}