zoukankan      html  css  js  c++  java
  • SZU:A26 Anagram

    Judge Info

    • Memory Limit: 32768KB
    • Case Time Limit: 10000MS
    • Time Limit: 10000MS
    • Judger: Normal

    Description

    An anagram is formed by rearranging the letters of a word. You are given a string, please find out if it is an anagram of a word or not. No word will have have more than 50 characters.

    Input

    The input will consist of a word on one line. The following line contains a number, n (1 leq n leq 30,000), of strings to be tested.

    Output

    For each test string, if the test string is identical to the source string, output 'IDENTICAL', if it is an anagram, output 'ANAGRAM' otherwise output 'NOT AN ANAGRAM', in a single line.

    Sample Input

    cares
    5
    scare
    races
    cares
    another
    acres
    

    Sample Output

    ANAGRAM
    ANAGRAM
    IDENTICAL
    NOT AN ANAGRAM
    ANAGRAM

    解题思路:字符串数组排序,但是我的方法并不好,只是勉强解出来而已。不过学会了使用qsort函数。

     1 #include <stdio.h>
     2 #include <string.h>
     3 char A[52];
     4 char B[52];
     5 char C[52];
     6 
     7 void swap(char *a,char *b){
     8     char t;
     9     t=*a;
    10     *a=*b;
    11     *b=t;
    12 }
    13 
    14 int main() {
    15     scanf("%s",A);
    16     int n,flag,i,j;
    17     scanf("%d",&n);
    18     for (i=0;i<strlen(A);++i){
    19         C[i]=A[i];
    20     } 
    21     while (n--) {
    22 
    23         scanf("%s",B);
    24         flag=1;
    25         for (i=0;i<strlen(A);++i) {
    26             if(A[i]!=B[i])
    27                 flag=0;
    28         }
    29         if(flag==1){printf("IDENTICAL
    "); continue;}
    30         for (i=0;i<strlen(C)-1;++i) {
    31             for (j=i+1;j<strlen(C);++j) {
    32                 if(B[i]>B[j])
    33                     swap(&B[i],&B[j]);
    34                 if(C[i]>C[j])
    35                     swap(&C[i],&C[j]);
    36             }
    37         }
    38         
    39         for (i=0;i<strlen(A);++i) {
    40             if(C[i]!=B[i])
    41                 flag=2;
    42         }
    43         if(flag==0){printf("ANAGRAM
    "); continue;}
    44         else printf("NOT AN ANAGRAM
    ");
    45     }
    46 }

    大神解法:

     1 #include<stdio.h>
     2 #include<stdlib.h>
     3 #include<string.h>
     4  
     5 char S[55];
     6  
     7 int cmp(const void *a,const void *b)
     8 {
     9     return *(char *)a-*(char *)b;
    10 }
    11  
    12 int main()
    13 {
    14     int n,i,len1,len2;
    15     char str[55],temp[55];
    16     scanf("%s",S);
    17     strcpy(temp,S);
    18     len1=strlen(S);
    19     qsort(S,len1,sizeof(char),cmp);
    20     scanf("%d",&n);
    21     for(i=0;i<n;i++)
    22     {
    23         memset(str,0,sizeof(str));
    24         scanf("%s",str);
    25         len2=strlen(str);
    26         if(len2!=len1)
    27         {
    28             printf("NOT AN ANAGRAM
    ");
    29             continue;
    30         }
    31         if(0==strcmp(str,temp))
    32         {
    33             printf("IDENTICAL
    ");
    34             continue;
    35         }
    36         else
    37         {
    38             qsort(str,len2,sizeof(char),cmp);
    39             if(0==strcmp(S,str))
    40                 printf("ANAGRAM
    ");
    41             else
    42                 printf("NOT AN ANAGRAM
    ");
    43         }
    44     }
    45     return 0;
    46 }
  • 相关阅读:
    DDL、DML、TCL
    SQL简介
    Java基础--常用API--IO流相关API
    Java基础--多线程
    Java基础--常用API--集合类相关API
    Java基础--常用API--日期相关API
    Java基础--常用API--字符串相关API
    Java基础--正则表达式的规则
    Java基础--常用API--java.lang.Object
    26、springboot——整合JPA
  • 原文地址:https://www.cnblogs.com/firstrate/p/3189446.html
Copyright © 2011-2022 走看看