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  • LightOJ

    Time Limit: 2000MS   Memory Limit: 32768KB   64bit IO Format: %lld & %llu

    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
    取数注意循环终止条件 ;
    #include <cstdio>
    #include <cstring>
    int main()
    {
        int c[5], d[5];
        char a[40], b[40];    
        int t, Qt=1; scanf("%d", &t);
        while(t--)
        {
            scanf("%s%s", a, b);
            int indexc=0, sum=0;
            int lena=strlen(a);
            for(int i=0; i<=lena; i++)
            {
                if(a[i]=='.'||i==lena)
                {
                    c[indexc++]=sum; sum=0;
                }
                else
                    sum=sum*10+a[i]-'0';
            }
            
            int indexd=0, Q=1; sum=0;
            int lenb=strlen(b);
            for(int i=lenb-1; i>=-1; i--) 
            {
                if(b[i]=='.'||i==-1)
                {
                    d[indexd++]=sum;
                    Q=1; sum=0;
                }
                else
                {
                    sum=sum+(b[i]-'0') *Q;
                    Q *=2;
                }
            }
            
            int i; 
            for(i=0; i< 4; i++)
                if(c[i] != d[3-i])
                    break;
            printf("Case %d: ", Qt++);
            if(i==4)
                printf("Yes
    ");
            else
                printf("No
    ");
        }
    }                         
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  • 原文地址:https://www.cnblogs.com/soTired/p/5357072.html
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