The Famous Clock
The Famous Clock
时间限制:1000 ms | 内存限制:65535 KB
难度:1
- 描述
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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?
- 输入
- 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.
- 输出
- For each test case, display a single line containing a decimal number corresponding to the given Roman Numerals.
- 样例输入
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I II III IV V VI VII VIII IX X XI XII
- 样例输出
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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
我写的代码有点小长,但是很容易就理解了。 -
#include<stdio.h> int main() { char a[5]; int i=0; while(scanf("%s",&a)!=EOF) { i+=1; if(a[0]=='I'&&a[1]=='\0') printf("Case %d: 1\n",i); else if(a[0]=='I'&&a[1]=='I'&&a[2]=='\0') printf("Case %d: 2\n",i); else if(a[0]=='I'&&a[1]=='I'&&a[2]=='I'&&a[3]=='\0') printf("Case %d: 3\n",i); else if(a[0]=='I'&&a[1]=='V'&&a[2]=='\0') printf("Case %d: 4\n",i); else if(a[0]=='V'&&a[1]=='\0') printf("Case %d: 5\n",i); else if(a[0]=='V'&&a[1]=='I'&&a[2]=='\0') printf("Case %d: 6\n",i); else if(a[0]=='V'&&a[1]=='I'&&a[2]=='I'&&a[3]=='\0') printf("Case %d: 7\n",i); else if(a[0]=='V'&&a[1]=='I'&&a[2]=='I'&&a[3]=='I'&&a[4]=='\0') printf("Case %d: 8\n",i); else if(a[0]=='I'&&a[1]=='X'&&a[2]=='\0') printf("Case %d: 9\n",i); else if(a[0]=='X'&&a[1]=='\0') printf("Case %d: 10\n",i); else if(a[0]=='X'&&a[1]=='I'&&a[2]=='\0') printf("Case %d: 11\n",i); else if(a[0]=='X'&&a[1]=='I'&&a[2]=='I'&&a[3]=='\0') printf("Case %d: 12\n",i); } return 0; }