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, , 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。
#include <stdio.h>
#include <string.h>
char A[];
char B[];
char C[]; void swap(char *a,char *b){
char t;
t=*a;
*a=*b;
*b=t;
} int main() {
scanf("%s",A);
int n,flag,i,j;
scanf("%d",&n);
for (i=;i<strlen(A);++i){
C[i]=A[i];
}
while (n--) { scanf("%s",B);
flag=;
for (i=;i<strlen(A);++i) {
if(A[i]!=B[i])
flag=;
}
if(flag==){printf("IDENTICAL\n"); continue;}
for (i=;i<strlen(C)-;++i) {
for (j=i+;j<strlen(C);++j) {
if(B[i]>B[j])
swap(&B[i],&B[j]);
if(C[i]>C[j])
swap(&C[i],&C[j]);
}
} for (i=;i<strlen(A);++i) {
if(C[i]!=B[i])
flag=;
}
if(flag==){printf("ANAGRAM\n"); continue;}
else printf("NOT AN ANAGRAM\n");
}
}
大神解法：
#include<stdio.h>
#include<stdlib.h>
#include<string.h> char S[]; int cmp(const void *a,const void *b)
{
return *(char *)a-*(char *)b;
} int main()
{
int n,i,len1,len2;
char str[],temp[];
scanf("%s",S);
strcpy(temp,S);
len1=strlen(S);
qsort(S,len1,sizeof(char),cmp);
scanf("%d",&n);
for(i=;i<n;i++)
{
memset(str,,sizeof(str));
scanf("%s",str);
len2=strlen(str);
if(len2!=len1)
{
printf("NOT AN ANAGRAM\n");
continue;
}
if(==strcmp(str,temp))
{
printf("IDENTICAL\n");
continue;
}
else
{
qsort(str,len2,sizeof(char),cmp);
if(==strcmp(S,str))
printf("ANAGRAM\n");
else
printf("NOT AN ANAGRAM\n");
}
}
return ;
}