contest: Codeforces Round #297 (Div. 二), problem: (B) Pasha and String

contest: Codeforces Round #297 (Div. 2), problem: (B) Pasha and String

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 ai and reversed a piece of string (a segment) from position ai to position|s| - ai + 1. It is guaranteed that 2·ai ≤ |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 ai (1 ≤ ai; 2·ai ≤ |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 test(s)

input
abcdef
1
2

output
aedcbf

input
vwxyz
2
2 2

output
vwxyz

input
abcdef
3
1 2 3

output
fbdcea

转置与顺序无关，比如 1 2 3 4 5 和 5 4 3 2 1的转置顺序结果是一样的，所以先进行排序，删除成对的且相同的，同一部分转置两次相当于没转；

其实还可以简化，刚开始嫌麻烦就这样交了..于是超时了..刚好卡在最后一组数据上..

于是要进一步优化，例如一个长度为10的串，输入 5 4 3 2 2 1 先排序，删掉两个2，剩下 1 3 4 5

即 1-10,3-8,4-7,5-6  先看前两个 3-8转了两次 所以只需要转置 1，2,与9,10即可 同理 只要交换4与7即可

这样不仅运行次数减少了一半以上，而且每次执行的任务也大大减少了，于是只花了31ms

code = =

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn1=200000+100,maxn2=100000+100;
char str[maxn1];     ///全局变量！！
int a[maxn2];
///char s[6];

void reve(int p,int q)   ///第一步优化,不用把数组作为参数加进来了
{
   for(int i=p;i<=(p+q)/2;i++)
   {
      char c=str[i];
      str[i]=str[p+q-i];
      str[p+q-i]=c;
   }
}

void change(int a,int b,int c,int d)    ///第二步优化
{
   for(int i=a;i<=b;i++)
     {
       char c=str[i];
       str[i]=str[a+d-i];
       str[a+d-i]=c;
     }
}


int main()
{

   /// s[0]='h';s[1]='e';s[2]='l';s[3]='l';s[4]='o',s[5]='\0';
  ///  reve(0,4);
   /// printf("%s\n",s);
    while(~scanf("%s",str))
    {
        int len=strlen(str);
       /// printf("%s\n",str);

        int m;
        scanf("%d",&m);   ///刚开始又忘了输入m了 = = 真心无奈...
        for(int i=0;i<m;i++)
            scanf("%d",a+i);
        sort(a,a+m);
       /// for(int i=0;i<m;i++)
          ///  printf("%d\n",a[i]);

        for(int i=0;i<m-1;i++)   ///i<m-1
            if(a[i]==a[i+1]&&a[i]!=0)  a[i]=0,a[i+1]=0;   ///a[i]!=0

        int flag=2;
        int k,t;
        for(int i=0;i<m;i++)
        {
            if(a[i])
            {
               if(flag==2) flag--,k=a[i]-1;   ///选出第一个没被删的数
               else if(flag==1) flag--,t=a[i]-1;    ///第二个没被删的数，这两个数一定不相等
            }

            if(!flag) change(k,t-1,len-t,len-k-1),flag=2;   ///优化2
        }
        if(flag==1) reve(k,len-1-k);   /// 是否存在落单的数


    ///    for(int i=0;i<m;i++)
           /// if(a[i]) reve(a[i]-1,len-a[i]);

        printf("%s\n",str);
    }
    return 0;
}

contest: Codeforces Round #297 (Div. 二), problem: (B) Pasha and String

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