描述
Mr. B, Mr. G and Mr. M are now in Warsaw, Poland, for the 2012’s ACM-ICPC World Finals 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. 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 represented in Roman Numerals. Mr. B cannot understand such kind of numbers. Can you translate for him?
B先生、G先生和M先生現在正在波蘭華沙參加2012年ACM - ICPC世界決賽。他們決定在比賽前每天進行5小時的訓練。此外，他們計劃每天10點開始訓練，因為世界決賽將會這樣做。華沙的風景如此迷人，B先生總是想在早飯后到外面散散步。不過，B先生必須在訓練開始前回去，否則他的隊友會很惱火。這里有個問題: B先生沒有手表。為了知道確切的時間，他在華沙買了一只新表，但那表上的所有數字都用羅馬數字表示。先生不能理解這種數字。你能替他翻譯嗎？

輸入
Each test case contains a single line indicating a Roman Numerals that to be translated. All the numbers can be found on clocks. That is, each number in the input represents an integer between 1 and 12. Roman Numerals are expressed by strings consisting of uppercase ‘I’, ‘V’ and ‘X’. See the sample input for further information.
每個測試用例包含一行，指示要翻譯的羅馬數字。所有的數字都可以在時鐘上找到。也就是說，輸入中的每個數字表示1到12之間的整數。羅馬數字由大寫字母“I”、“V”和“X”組成的字符串表示。有關詳細信息，請參閱示例輸入。
輸出
For each test case, display a single line containing a decimal number corresponding to the given Roman Numerals.
對于每個測試用例，顯示一行包含對應于給定羅馬數字的十進制數。
樣例輸入
I
II
III
IV
V
VI
VII
VIII
IX
X
XI
XII樣例輸出
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

代碼如下

#include< stdio.h>
#include< string.h>
int main()
{
char a[100];//定義一個字符數組；
int x,y=1;//這里的y負責輸出Case后面的那個數；
while(scanf(“%s”,&a)!=EOF)//然后依次比較這12個羅馬數字；
{
if(strcmp(a,”I”)==0)
{
x=1;//這里的x負責羅馬數字的翻譯，即將羅馬數字翻譯成阿拉伯數字輸出出來；
}
else if(strcmp(a,”II”)==0)
{
x=2;
}
else if(strcmp(a,”III”)==0)
{
x=3;
}
else if(strcmp(a,”IV”)==0)
{
x=4;
}
else if(strcmp(a,”V”)==0)
{
x=5;
}
else if(strcmp(a,”VI”)==0)
{
x=6;
}
else if(strcmp(a,”VII”)==0)
{
x=7;
}
else if(strcmp(a,”VIII”)==0)
{
x=8;
}
else if(strcmp(a,”IX”)==0)
{
x=9;
}
else if(strcmp(a,”X”)==0)
{
x=10;
}
else if(strcmp(a,”XI”)==0)
{
x=11;
}
else
{
x=12;
}
printf(“Case %d: %d\n”,y++,x);//最后輸出y++（由于每次輸出都要+1，y為第一次的輸出）；
}
return 0;
}

總結

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