题目描述
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
Constraints
Each character in s is a lowercase English letter.
1≤|s|≤105
Constraints
Each character in s is a lowercase English letter.
1≤|s|≤105
输入
The input is given from Standard Input in the following format:
s
s
输出
Print the string obtained by concatenating all the characters in the odd-numbered positions.
样例输入
atcoder
样例输出
acdr
提示
Extract the first character a, the third character c, the fifth character d and the seventh character r to obtain acdr.
题意:求奇数位置的字符
今天傻逼了。。。。。。。
写此博客以示警告
strlen()这个函数的时间复杂度是0(n),如果for(int i = 0; i < strlen(str); i++)这样操作,时间复杂度会是 0(n*n),但string 类型调用length()方法的时间复杂度是0(1)
AC code:
#include <bits/stdc++.h> using namespace std; const int N =1e6+10; char str[N]; int main() { scanf("%s",str); int krn=strlen(str); for(int i = 0;i < krn;i++) if(!(i&1)) printf("%c",str[i]); printf(" "); return 0; }