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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)
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