Codeforces #297 (Div. 2) B. Pasha and String （贪心_codeforces 293div2 pasha and pipe-优快云博客

本文链接：https://blog.youkuaiyun.com/wang2332/article/details/64615209

本文介绍了一个关于字符串翻转的问题及其解决方案。Pasha在生日那天收到了一个由小写字母组成的字符串，通过一系列的翻转操作后，最终得到一个新的字符串。文章详细解释了如何通过贪心算法快速求解这一问题，并提供了完整的C++代码实现。

B. Pasha and String

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Pasha got a very beautiful string s for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |s| from left to right, where |s| is the length of the given string.

Pasha didn’t like his present very much so he decided to change it. After his birthday Pasha spent m days performing the following transformations on his string — each day he chose integer a_i and reversed a piece of string (a segment) from position a_i to position |s| - a_i + 1. It is guaranteed that 2·a_i ≤ |s|.

You face the following task: determine what Pasha’s string will look like after m days.

Input

The first line of the input contains Pasha’s string s of length from 2 to 2·10⁵ characters, consisting of lowercase Latin letters.

The second line contains a single integer m (1 ≤ m ≤ 10⁵) — the number of days when Pasha changed his string.

The third line contains m space-separated elements a_i (1 ≤ a_i; 2·a_i ≤ |s|) — the position from which Pasha started transforming the string on the i-th day.

Output

In the first line of the output print what Pasha’s string s will look like after m days.

Examples

Input

abcdef
1
2

Output

aedcbf

Input

vwxyz
2
2 2

Output

vwxyz

Input

abcdef
3
1 2 3

Output

fbdcea

推了很长时间的规律，贪心问题

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 200000+100
char str[N];
int vis[N];

int main()
{
    int n, x;
    scanf("%s",str);
    scanf("%d",&n);
    for(int i = 0;i < n; i++) {
        scanf("%d",&x);
        vis[x]++;
    }
    int len = (int)strlen(str);
    int sum = 0;
    for(int i = 0;i < len/2; i++) {
        sum += vis[i+1];
        if(sum&1) swap(str[i], str[len-1-i]);
    }
    printf("%s\n",str);
return 0;
}