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)。

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#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; }