ZOJ 1068 P,MTHBGWB

原题链接

题目大意:给定一个字符串,先用Morse Code编码,把编码倒序,再解码成字符串。现给定处理后的字符串,求原始信息。

解法:用C++String类的函数。每次读入一个字符,就在string后面接上,并且保存字符对应morse码的长度。这里要声明一下,真正的morse码也是有句号、问号这些常用标点符号的,但不是题目里给出的那种形式。比如,句号是“.-.-.-”。

参考代码:

#include<iostream>
#include<string>
using namespace std;

string code[30]={".-","-...","-.-.","-..",".","..-.","--.","....","..",".---",
				"-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-",
				"...-",".--","-..-","-.--","--..","..--",".-.-","---.","----"};


int main(){
	int i,j,k,len,n,num[100],cases=0;
	string str,ss;

	cin>>n;
	while(n--){
		cases++;
		cin>>str;
		len=str.length();
		i=0;
		string morse="";
		while(i<len){
			if(str[i]>64&&str[i]<91){
				morse.append(code[str[i]-'A']);
				num[len-1-i]=code[str[i]-'A'].size();
			}				
			else{
				switch(str[i]){
				case '_': morse.append("..--");break;
				case ',': morse.append(".-.-");break;
				case '.': morse.append("---.");break;
				case '?': morse.append("----");break;
				}
				num[len-1-i]=4;
			}
			i++;
		}
		i=j=0;
		while(i<len){
			k=0;
			ss=morse.substr(j,num[i]);
			j+=num[i];
			if(ss=="..--"){str[i++]='_';continue;}
			if(ss==".-.-"){str[i++]=',';continue;}
			if(ss=="---."){str[i++]='.';continue;}
			if(ss=="----"){str[i++]='?';continue;}
			while(k<26){
				if(ss.compare(code[k])==0){
					str[i++]=k+'A';
					break;
				}
				k++;
			}
		}
		cout<<cases<<": "<<str<<endl;


	}
		


	return 0;
}
原文地址:https://www.cnblogs.com/naive/p/3568759.html