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

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本文介绍了一道关于字符串翻转的算法题——Pasha和字符串。文章详细阐述了如何通过统计每位数字的翻转次数来高效解决该问题，并提供了一个C++实现示例。

题目链接：come on！！

题目思路：这个题目最开始我直接统计每一位是奇数还是偶数，但是后来一直tle，后来想其实这根本没有起到优化的作用，后来发现其实其实我的思路是对的，但是也不对，其实应该统计每一位交换是奇数还是偶数次（采用抑或即可），但是对于那些，没有出现的数位，其实在被动的被交换，所以当一位是奇数时标志改变，后面的都要进行交换，然后扫描一半的字符长度即可！！！

题目：

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

代码：

#include<cstdio>
#include<cstring>

const int maxn=200000+10;
char s[maxn];

int a[maxn],m,b[maxn];

int main()
{
    while(~scanf("%s",s+1))
    {
        scanf("%d",&m);
        memset(b,0,sizeof(b));
        int l=strlen(s+1);
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&a[i]);
            b[a[i]]^=1;
        }
        bool flag=false;
        for(int i=1;i<=l/2;i++)
        {
            if(b[i])  flag^=true;
            if(flag)
            {
                char c=s[i];
                s[i]=s[l-i+1];
                s[l-i+1]=c;
            }
        }
        printf("%s",s+1);
    }
    return 0;
}