lightoj--1354-- IP Checking（水题）

IP Checking

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Submit Status

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

Input starts with an integer T (≤ 100), denoting the number of test cases.

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 the case number and "Yes" if they are same, otherwise print "No".

Sample Input

2

192.168.0.100

11000000.10101000.00000000.11001000

65.254.63.122

01000001.11111110.00111111.01111010

Sample Output

Case 1: No

Case 2: Yes

Source

Problem Setter: Jane Alam Jan

Submit Status

#include<stdio.h>
#include<string.h>
#include<math.h>
int a[4];
char str[50];
int main()
{
	int t;
	scanf("%d",&t);
	int Case=1;
	while(t--)
	{
		memset(str,'',sizeof(str));
		memset(a,0,sizeof(a));
		scanf("%d.%d.%d.%d",&a[0],&a[1],&a[2],&a[3]);
		str[0]='.';
//啊，真是打脸，我从最后一位判断，但是我竟然忘了，第一位不是‘.’，wa了n次 
		scanf("%s",str+1);
		int j=3,flog=0,sum=0;
		int k=0;
		for(int i=strlen(str)-1;i>=0;i--)
		{
			if(str[i]=='1')
			sum+=pow(2,k);
			else if(str[i]=='.') 
			{
				if(sum!=a[j])
				{
					flog=1;break;
				}
				sum=0;k=0;j--;
				continue;
			}
			k++;
		}
		printf("Case %d: ",Case++);
		if(flog) printf("No
");
		else printf("Yes
");
	}
	return 0;
}