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
#include
#include
#include
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