• lightoj-1354


    1354 - IP Checking
    PDF (English) Statistics Forum
    Time Limit: 2 second(s) Memory Limit: 32 MB
    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
    Output for Sample Input
    2
    192.168.0.100
    11000000.10101000.00000000.11001000
    65.254.63.122
    01000001.11111110.00111111.01111010
    Case 1: No
    Case 2: Yes

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<algorithm>
     4 using namespace std;
     5 
     6 int main(){
     7     
     8     int T,a[4],a1[4];
     9     
    10     scanf("%d",&T);
    11     for(int t=1;t<=T;t++){
    12         
    13         scanf("%d.%d.%d.%d",&a[0],&a[1],&a[2],&a[3]);
    14         scanf("%d.%d.%d.%d",&a1[0],&a1[1],&a1[2],&a1[3]);
    15         int flag = 0;
    16         for(int j=0;j<4&&flag==0;j++)
    17             for(int i=0;i<8;i++){
    18                 if((a[j]&1)!=a1[j]%10){
    19                     flag = 1;
    20                     break;
    21                 }
    22                 a[j]/=2,a1[j]/=10;
    23             }
    24 
    25         printf("Case %d: ",t);
    26         printf(flag==0?"Yes
    ":"No
    ");
    27     }
    28     
    29     return 0;
    30 } 
    31 //未优化
    32 //int main(){
    33 //    
    34 //    int T,a,b,c,d,a1,b1,c1,d1;
    35 //    
    36 //    scanf("%d",&T);
    37 //    for(int t=1;t<=T;t++){
    38 //        
    39 //        scanf("%d.%d.%d.%d",&a,&b,&c,&d);
    40 //        scanf("%d.%d.%d.%d",&a1,&b1,&c1,&d1);
    41 //        int flag = 0;
    42 //        
    43 //        for(int i=0;i<8;i++){
    44 //            if((a&1)!=a1%10){
    45 //                flag = 1;
    46 //                break;
    47 //            }
    48 //            a/=2,a1/=10;
    49 //        }
    50 //        if(flag==0){
    51 //            for(int i=0;i<8;i++){
    52 //                if((b&1)!=b1%10){
    53 //                    flag = 1;
    54 //                    break;
    55 //                }
    56 //                b/=2,b1/=10;
    57 //            }
    58 //        }
    59 //        if(flag==0){
    60 //            for(int i=0;i<8;i++){
    61 //                if((c&1)!=c1%10){
    62 //                    flag = 1;
    63 //                    break;
    64 //                }
    65 //                c/=2,c1/=10;
    66 //            }
    67 //        }
    68 //        if(flag==0){
    69 //            for(int i=0;i<8;i++){
    70 //                if((d&1)!=d1%10){
    71 //                    flag = 1;
    72 //                    break;
    73 //                }
    74 //                d/=2,d1/=10;
    75 //            }
    76 //        }
    77 //        printf("Case %d: ",t);
    78 //        printf(flag==0?"Yes
    ":"No
    ");
    79 //    }
    80 //    
    81 //    return 0;
    82 //} 
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  • 原文地址:https://www.cnblogs.com/yuanshixingdan/p/5564038.html
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