zoukankan      html  css  js  c++  java
  • Codeforces Round #528-A. Right-Left Cipher(字符串模拟)

    time limit per test

    1 second

    memory limit per test

    256 megabytes

    input

    standard input

    output

    standard output

    Polycarp loves ciphers. He has invented his own cipher called Right-Left.

    Right-Left cipher is used for strings. To encrypt the string s=s1s2…sns=s1s2…sn Polycarp uses the following algorithm:

    • he writes down s1s1,
    • he appends the current word with s2s2 (i.e. writes down s2s2 to the right of the current result),
    • he prepends the current word with s3s3 (i.e. writes down s3s3 to the left of the current result),
    • he appends the current word with s4s4 (i.e. writes down s4s4 to the right of the current result),
    • he prepends the current word with s5s5 (i.e. writes down s5s5 to the left of the current result),
    • and so on for each position until the end of ss.

    For example, if ss="techno" the process is: "t" →→ "te" →→ "cte" →→ "cteh" →→ "ncteh" →→ "ncteho". So the encrypted ss="techno" is "ncteho".

    Given string tt — the result of encryption of some string ss. Your task is to decrypt it, i.e. find the string ss.

    Input

    The only line of the input contains tt — the result of encryption of some string ss. It contains only lowercase Latin letters. The length of tt is between 11 and 5050, inclusive.

    Output

    Print such string ss that after encryption it equals tt.

    Examples

    input

    Copy

    ncteho
    

    output

    Copy

    techno
    

    input

    Copy

    erfdcoeocs
    

    output

    Copy

    codeforces

    input

    Copy

    z
    

    output

    Copy

    z

    分一下字符串的奇偶模拟即可,注意要用char 不要用string

    代码:

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    
    using namespace std;
    
    int main() {
    
    	char str[10005],str1[10005];
    	scanf("%s",str);
    	int len=strlen(str);
    	int s=0;
    	if(len%2==0) {
    
    		for(int t=len/2-1; t>=0; t--) {
    			str1[s++]=str[t];
    			str1[s++]=str[len-1-t];
    		}
    	} else {
    		str1[s++]=str[len/2];
    		for(int t=len/2-1; t>=0; t--) {
    			str1[s++]=str[len-1-t];
    			str1[s++]=str[t];
    		}
    	}
    	for(int t=0; t<s; t++) {
    		cout<<str1[t];
    	}
    	return 0;
    }
  • 相关阅读:
    地址栏传值 JS取值方法
    定位导航 制作
    验证码
    图片水印
    AJAX 三级联动
    javascript 和Jquery 互转
    Jquery 事件 DOM操作
    Jquery 基础
    软件工程中的形式化方法读后感
    软件工程理论、方法与实践 需求工程读后感
  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10781868.html
Copyright © 2011-2022 走看看