UVA - 401 Palindromes
来源:互联网 发布:淘宝直通车养词要多久? 编辑:程序博客网 时间:2024/06/03 04:10
题目大意:判断是否回文、镜面。回文即从左往右与从右往左相同。镜面根据所提供的表格,可能存在非法字符。一开始理解错题意,以为不存在非法字符就是镜面,实际上是从左往右根据表格将全部字符化为镜像后,要与从右往左相同。
解题思路:回文简单不赘述。用数组把镜像按顺序存储,字母和数字要分开,判断的时候读取对应镜像与从右数相同位置的字符比较。范围是 (l+1)/2 不是 l/2,卡了好久QwQ
#include<iostream>#include<cstdio>#include<string.h>using namespace std;char str[1000005];char zimu[]="A 3 HIL JM O 2TUVWXY5";char shuzi[]="1SE Z 8 ";int main() { memset (str, '0', sizeof(str)); while (scanf("%s", str) != EOF){ int l = strlen(str); int pa = 1, mi = 1; for (int i = 0; i < (l+1)/2; i++){ if (str[i] != str[l-i-1]) pa = 0; if (str[i] >= 'A' && str[i] <= 'Z' && zimu[str[i]-'A'] != str[l-i-1]) mi = 0; if (str[i] >= '1' && str[i] <= '9' && shuzi[str[i]-'1'] != str[l-i-1]) mi = 0; } printf("%s", str); if (pa+mi == 0) printf(" -- is not a palindrome.\n"); if (pa+mi == 1) printf(pa?" -- is a regular palindrome.\n":" -- is a mirrored string.\n"); if (pa+mi == 2) printf(" -- is a mirrored palindrome.\n"); printf("\n"); }return 0;}
0 0
- UVa 401 Palindromes
- UVa OJ 401-Palindromes
- UVa:401 - Palindromes
- uva 401 Palindromes //字符串
- uva 401 - Palindromes
- UVa 401 - Palindromes
- UVa 401 - Palindromes
- UVa 401 - Palindromes
- uva-401 - Palindromes
- UVa 401 Palindromes
- uva 401 Palindromes
- uva 401 Palindromes
- UVA 401 - Palindromes
- UVa 401 - Palindromes
- UVA 题目401 - Palindromes
- UVA 401 Palindromes
- UVa 401: Palindromes
- UVA 401 Palindromes
- Source Insight 常用设置和快捷键大全
- 升级款E4418CORE-V1C 最强最小工业级核心模块 横空出世
- ASP.NET-----验证控件
- 代理模式
- JAVA中线程池的整理
- UVA - 401 Palindromes
- iOS - 报错 Warning: Attempt to present <xx: xx> on <xx: xx> whose view is not in the window hierarchy!
- 开源MySQL数据仓库解决方案:Infobright
- [Windows] MBN send SMS pdu
- 欢迎使用CSDN-markdown编辑器
- BroadcastReceiver 应用之apk自启动
- Hibernate Mapping 1 : OneToOne
- Python里类变量和实例变量的区别
- 极光推送的简单实现方法