Codeforces Round #297 (Div. 2)B. Pasha and String 前缀和-白红宇

Codeforces Round #297 (Div. 2)B. Pasha and String 前缀和

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发布时间：2019-06-12

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B. Pasha and String

Time Limit: 2 Sec Memory Limit: 256 MB

Submit: xxx Solved: 2xx

题目连接

http://codeforces.com/contest/525/problem/B

Description

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.

Sample Input

Input

abcdef 1 2

Input

vwxyz 2 2 2

Input

abcdef 3 1 2 3

Sample Output

Output

aedcbf

Output

vwxyz

Output

fbdcea

HINT

题意：

给你一个字符串，然后给你一个数字i，然后i到s.size()-i直接的位置全部翻转，然后问你m次操作之后，是什么样子

题解：

首先，只有翻转奇数次和翻转偶数次两种情况，我们用一个类似前缀和的操作，就可以处理出这个位置究竟翻转了多少次，然后输出就好啦

~\(≧▽≦)/~啦啦啦，这道题完啦

代码：

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                 typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 400010#define mod 10007#define eps 1e-9//const int inf=0x7fffffff; //无限大const int inf=0x3f3f3f3f;/**///**************************************************************************************inline ll read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}int flag[maxn];int main(){ string s; cin>>s; int n; cin>>n; for(int i=0;i

转载于:https://www.cnblogs.com/qscqesze/p/4371790.html

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