题目描述：
IP Checking
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 128000/64000 KB (Java/Others)
Problem Description
An IP address is a 32 bit address formatted in the following way:
a.b.c.d
where a, b, c, d are integers each ranging from 0 to 255. Now you are given two IP addresses, first one in decimal form and second one in binary form, your task is to find if they are same or not.
Input
Each case starts with two lines.
First line contains an IP address in decimal form, and second line contains an IP address in binary form. In binary form, each of the four parts contains 8 digits. Assume that the given addresses are valid.
Output
For each case, print "Yes" if they are same, otherwise print "No".
Sample Input
192.168.0.100 11000000.10101000.00000000.11001000 65.254.63.122 01000001.11111110.00111111.01111010
Sample Output
No Yes
死改死改死改！！！原来"No" 和 "Yes" 输出变成 "NO" 和 "YES" 了，而且还是问asdfgg的，瘀死了 = =！！记录下来是警醒自己以后不能再犯这种错误了！！！
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 using namespace std; 6 7 const int maxn = 50; 8 char s2[maxn]; 9 int s[5]; 10 11 int main() 12 { 13 char ch; 14 int t1, t2, t3, t4, l1, l2, l, i, j, p, tmp; 15 while (scanf("%d%c%d%c%d%c%d", &t1, &ch, &t2, &ch, &t3, &ch, &t4) != EOF) 16 { 17 scanf("%s", s2); 18 l2 = strlen(s2); 19 l1 = 7, l = tmp = 0; 20 for (i = 0; i < l2; i++) 21 { 22 if (s2[i] != '.') 23 { 24 p = 1; 25 for (j = 1; j <= l1; j++) 26 p *= 2; 27 tmp += p * (s2[i]-'0'); 28 l1--; 29 } 30 else 31 { 32 s[l++] = tmp; 33 tmp = 0; 34 l1 = 7; 35 } 36 } 37 s[l++] = tmp; 38 if (s[0] == t1 && s[1] == t2 && s[2] == t3 && s[3] == t4) 39 printf("Yes "); 40 else 41 printf("No "); 42 } 43 return 0; 44 }