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PAT-1100(Mars Numbers)

  题目见这里

   题目并不难,不过一开始我没能理清题意,参考了下这里,明白题目实际是考进制转换(13进制),外加一个映射(hash),这当然比较简单啦!需要注意的是,样例中tam(Mars Number)----->13(Earth),由此可知,以13的整倍数(Earth)出现时,不带‘tret’(earth:0)

#include <cstdio>#include <iostream>#include <map>#include <cstring>using namespace std;struct ptrCmp{	bool operator()(const char *s1, const char *s2) const{		return strcmp(s1,s2)<0;	}};int main(){	const char hash[25][5]={"tret","jan","feb","mar","apr","may","jun","jly","aug","sep","oct","nov","dec",	 "tam","hel","maa","huh","tou","kes","hei","elo","syy","lok","mer","jou"};	map<const char*,int,ptrCmp>hash1;	int i;	for(i=0;i<25;i++)		hash1.insert(pair<const char*,int>(hash[i],i));  //  freopen("Data.txt","r",stdin);	char s[5],s1[5],c;	int n;	scanf("%d",&n);	getchar();	while(n--){		scanf("%s",s);		scanf("%c",&c);		if(c==‘ ‘ || (s[0]>=‘a‘ && s[0]<=‘z‘)){			if(c==‘ ‘) scanf("%s",s1);			if(c==‘\n‘){				int index = hash1[s];				if(index<13) printf("%d",index);				else printf("%d",(index-12)*13);			}			else printf("%d",(hash1[s]-12)*13+hash1[s1]);			printf("\n");		}		else{			int key,k;			key=0,k=0;			while(s[k]) key = 10*key+s[k]-‘0‘, k ++;			if(key<13) printf("%s\n",hash[key]);			else{				printf("%s",hash[key/13+12]);					if(key%13) printf(" %s",hash[key%13]); //不带‘tret‘ 				printf("\n");			}		}	}	return 0;}

 

PAT-1100(Mars Numbers)