1354 - IP Checking PDF (English) Statistics ForumTime Limit: 2 second(s) Memory Limit: 32 MBAn 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.
InputInput 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.
OutputFor each case, print the case number and "Yes" if they are same, otherwise print "No".
Sample InputOutput for Sample Input2192.168.0.10011000000.10101000.00000000.1100100065.254.63.12201000001.11111110.00111111.01111010Case 1: NoCase 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\n":"No\n"); 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){ 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){ 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){ 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){ 71 // flag = 1; 72 // break; 73 // } 74 // d/=2,d1/=10; 75 // } 76 // } 77 // printf("Case %d: ",t); 78 // printf(flag==0?"Yes\n":"No\n"); 79 // } 80 // 81 // return 0; 82 //}
转载于:https://www.cnblogs.com/yuanshixingdan/p/5564038.html
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