PAT甲级1100 Mars Numbers（模拟）

最新推荐文章于 2025-07-18 14:48:14 发布

threeh20

最新推荐文章于 2025-07-18 14:48:14 发布

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本文解析了1100 MarsNumbers题目，详细介绍了如何实现地球数字和火星数字之间的转换，包括处理火星文特有的计数方式和特殊情况。

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1100 Mars Numbers （20 分）

People on Mars count their numbers with base 13:

Zero on Earth is called "tret" on Mars.
The numbers 1 to 12 on Earth is called "jan, feb, mar, apr, may, jun, jly, aug, sep, oct, nov, dec" on Mars, respectively.
For the next higher digit, Mars people name the 12 numbers as "tam, hel, maa, huh, tou, kes, hei, elo, syy, lok, mer, jou", respectively.

For examples, the number 29 on Earth is called "hel mar" on Mars; and "elo nov" on Mars corresponds to 115 on Earth. In order to help communication between people from these two planets, you are supposed to write a program for mutual translation between Earth and Mars number systems.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<100). Then N lines follow, each contains a number in [0, 169), given either in the form of an Earth number, or that of Mars.

Output Specification:

For each number, print in a line the corresponding number in the other language.

Sample Input:

4
29
5
elo nov
tam

Sample Output:

hel mar
may
115
13

题意就是翻译语言，按照题意不难

但是有坑，就是火星文里有tret的话会省略，就像样例一样tam 而不是tam tret。

#include<iostream>
#include<cstdio>
#include<cstring> 
using namespace std;
char s1[14][4]={"tam","hel","maa","huh","tou","kes","hei","elo","syy","lok","mer","jou"};
char s2[14][5]={"tret","jan","feb","mar","apr","may","jun","jly","aug","sep","oct","nov","dec"}; 
int len;
void tran_to_mars(char a[]){
	char c[4][4];
	int b,t=1,sum=0;
	len=strlen(a);
	for(int i=len-1;i>=0;i--){
		b=a[i]-'0';
		b*=t;
		t*=10;
		sum+=b;
	}
	int ge=sum%13;
	int shi=sum/13;
	if(shi==0){
		printf("%s\n",s2[ge]);
	}else{
		if(ge!=0){
			printf("%s %s\n",s1[shi-1],s2[ge]);
		}else{
			printf("%s\n",s1[shi-1]);
		}
	}
}
void tran_to_earth(char a[]){
	int t=0;
	len=strlen(a);
	if(len==4)
	{
		cout<<"0"<<endl;
	}
	else if(len==3)
	{
		for(int i=0;i<12;i++)
		{
			if(a[0]==s2[i+1][0]&&a[1]==s2[i+1][1]&&a[2]==s2[i+1][2])
			{
				t=i+1;
			}
		}
		for(int i=0;i<12;i++)
		{
			if(a[0]==s1[i][0]&&a[1]==s1[i][1]&&a[2]==s1[i][2])
			{
				t=(i+1)*13;
			}
		}
		cout<<t<<endl;
	}
	else if(len==7)
	{
		for(int i=0;i<12;i++)
		{
			if(a[0]==s1[i][0]&&a[1]==s1[i][1]&&a[2]==s1[i][2])
			{
				t+=(i+1)*13;
			}
		}
		for(int i=0;i<12;i++)
		{
			if(a[4]==s2[i+1][0]&&a[5]==s2[i+1][1]&&a[6]==s2[i+1][2])
			{
				t+=i+1;
			}
		}
		cout<<t<<endl;
	}
}
int main(){
	int n;
	char a[11];
	cin>>n;
	getchar();
	while(n--){
		fgets(a,11,stdin);
		len=strlen(a);
		a[len-1]='\0';
		//scanf("%s",a);
		if(a[0]>='0'&&a[0]<='9'){
			tran_to_mars(a);
		}else{
			tran_to_earth(a);
		}
	}
	return 0;
}