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,
, 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 ;
}
SZU:A26 Anagram的更多相关文章
- [LeetCode] Valid Anagram 验证变位词
Given two strings s and t, write a function to determine if t is an anagram of s. For example, s = & ...
- Leetcode Valid Anagram
Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = &q ...
- LeetCode 242 Valid Anagram
Problem: Given two strings s and t, write a function to determine if t is an anagram of s. For examp ...
- 【09_242】Valid Anagram
Valid Anagram My Submissions Question Total Accepted: 43694 Total Submissions: 111615 Difficulty: Ea ...
- 【leetcode❤python】242. Valid Anagram
class Solution(object): def isAnagram(self, s, t): if sorted(list(s.lower()))==sorted(list ...
- 242. Valid Anagram
Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = &q ...
- (easy)LeetCode 242.Valid Anagram
Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = &q ...
- 【LeetCode】242 - Valid Anagram
Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = &q ...
- Java [Leetcode 242]Valid Anagram
题目描述: Given two strings s and t, write a function to determine if t is an anagram of s. For example, ...
随机推荐
- iOS 8中CLLocationManager及MKMapView showUserLocation失败的解决的方法
用XCode 6编译的原来XCode 5.1.1写的程序时,发现原来写的CLLocationManager定位的代码以及MKmapView的showUserLocation失效.查了一下,XCode ...
- java基金会成立
在java在,数据收集的操作,应使用非常.最近看了零星收集的小知识,在这里,一点点排序. 它基本上是四个常用的类操作点总结集合. 首先.集合大致分为两个方向.一种是普通的集合类型,通过接口collec ...
- Python开发一个csv比较功能相关知识点汇总及demo
Python 2.7 csv.reader(csvfile, dialect='excel', **fmtparams)的一个坑:csvfile被csv.reader生成的iterator,在遍历每二 ...
- 基于Android的ELF PLT/GOT符号和重定向过程ELF Hook实现(by 低端农业代码 2014.10.27)
介绍 技术原因写这篇文章,有两种: 一个是在大多数在线叙述性说明发现PLT/GOT第二十符号重定向过程定向x86的,例<Redirecting functions in shared ELF l ...
- Java数据结构与算法(13) - ch06递归(归并排序)
时间为O(N*logN). 归并排序的一个缺点是它需要在存储器中有另一个大小等于被排序的数据项数目的数组.归并两个有序的数组.利用递归,不断的将数组进行二分法排序,然后进行归并即可.
- oracle_体系结构图_逻辑结构图
1.oracle 的体系结构图 重要!!! 2.oracle的逻辑结构图
- Android 2.3.5源码 更新至android 4.4,能够下载,度娘网盘
Android 4.4源代码下载(linux合并) ==============================切割线结束========================= 旧版本号的能够使用115, ...
- 【百度地图API】如何自定义地图图层?实例:制作麻点图(自定义图层+热区)
原文:[百度地图API]如何自定义地图图层?实例:制作麻点图(自定义图层+热区) 摘要:自定义地图图层的用途十分广泛.常见的应用,比如制作魔兽地图和清华校园地图(使用切图工具即可轻松实现).今天我们来 ...
- ResultSet 转为 List或者JavaBean
一.将ResultSet结果集转换为List,其中每条记录信息保存为Map放到List中,方法如下: public static List<Map<String, Object>&g ...
- IIS7伪静态化URL Rewrite模块
原文 IIS7伪静态化URL Rewrite模块 在Win7安装了IIS7.5之后,搭建一些网站或者博客,但是IIS7.5本身没有URL Rewrite功能,也就是无法实现网址的伪静态化. 从网上找了 ...