• The Famous Clock


                                                                            The Famous Clock

                                          Time Limit:1000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

    Mr. B, Mr. G and Mr. M are now in Warsaw, Poland, for the 2012’s ACM-ICPC World Final Contest. They’ve decided to take a 5 hours training every day before the contest. Also, they plan to start training at 10:00 each day since the World Final Contest will do so. The scenery in Warsaw is so attractive that Mr. B would always like to take a walk outside for a while after breakfast. However, Mr. B have to go back before training starts, otherwise his teammates will be annoyed. But here is a problem: Mr. B does not have a watch. In order to know the exact time, he has bought a new watch in Warsaw, but all the numbers on that watch are presented with Roman numerals. Mr. B cannot understand such kind of numbers. Can you translate for him?

    Input

    Each test case contains a single line indicating a Roman number that to be translated. All the numbers can be found on clocks. That is, each numbers in the input represents an integer between 1 and 12. Roman numbers are expressed by strings consisting of uppercase I, Vand X. See the sample input for further information.

    Output

    For each test case, display a single line containing a decimal number corresponding to the given Roman number.

    Sample Input

    Input:
    I
    II
    III
    IV
    V
    VI
    VII
    VIII
    IX
    X
    XI
    XII
    
    Output:
    Case 1: 1
    Case 2: 2
    Case 3: 3
    Case 4: 4
    Case 5: 5
    Case 6: 6
    Case 7: 7
    Case 8: 8
    Case 9: 9
    Case 10: 10
    Case 11: 11
    Case 12: 12
    列举就好  注意列举的方法   “”符号的问题  等等
     
    代码:
    #include<stdio.h>
    #include<string.h>
    int main()
    {
        int c,t,i;
        char x[10];
        char y[20][10] = {"#","I","II","III","IV","V","VI","VII","VIII","IX","X","XI","XII"};
        t=1;
        while(~scanf("%s",x))
        {
            for(i = 1;i<=12;i++)
            {
                if(strcmp(x,y[i])==0)
                {
                    printf("Case %d: %d
    ",t,i);
                }
            }
            t++;
    
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lipching/p/3850598.html
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