Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 18418 | Accepted: 6759 |
Description
If the word is absent in the dictionary then it can be replaced by correct words (from the dictionary) that can be obtained by one of the following operations:
?
deleting of one letter from the word;
?replacing of one letter in the word with an arbitrary letter;
?
inserting of one arbitrary letter into the word.
Your task is to write the program that will find all possible replacements from the dictionary for every given word.
Input
The next part of the file contains all words that are to be checked. Each word occupies its own line. This part is also finished by the single character '#' on a separate line. There will be at most 50 words that are to be checked.
All words in the input file (words from the dictionary and words to be checked) consist only of small alphabetic characters and each one contains 15 characters at most.
Output
Sample Input
i is has have be my more contest me too if award # me aware m contest hav oo or i fi mre #
Sample Output
me is correct aware: award m: i my me contest is correct hav: has have oo: too or: i is correct fi: i mre: more me
Source
C++只是,G++过。。
。
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> using namespace std; struct node { char s[20]; int t; } dic[10010]; int ctt(char *str,char *ch) { int l1=strlen(str),l2=strlen(ch); int i,t=0; if(l1<l2) { for(i=0;i<l2;i++) { if(str[t]==ch[i]) t++; if(t==l1) return 1; } return 0; } else if(l1>l2) { for(i=0;i<l1;i++) { if(str[i]==ch[t]) t++; if(t==l2) return 1; } return 0; } else { for(i=0;i<l1;i++) { if(str[i]==ch[i]) t++; } if(t==l1-1) return 1; else return 0; } } int main() { int n=0,f=0,i; char str[20]; while(~scanf("%s",str)) { if(str[0]=='#')break; strcpy(dic[n].s,str); dic[n++].t=n; } //<<ctt("aware","award"); while(~scanf("%s",str)) { f=0; if(str[0]=='#')break; for(i=0; i<n; i++) { if(strcmp(str,dic[i].s)==0) { printf("%s is correct",str); f=1; break; } } if(!f) { printf("%s:",str); int l=strlen(str); for(i=0; i<n; i++) { if(strlen(dic[i].s)-l==1||strlen(dic[i].s)-l==-1||strlen(dic[i].s)-l==0) { if(ctt(str,dic[i].s)) printf(" %s"); } } } printf("\n"); } }