CF 525B(Pasha and String-贪心+找规律)

B. Pasha and String

time limit per test

memory limit per test

input

output

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|

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. It is guaranteed that 2·ai ≤ |s|.

m

Input

s of length from 2 to 2·105

m (1 ≤ m ≤ 105) —  the number of days when Pasha changed his string.

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

s will look like after m

Sample test(s)

input

abcdef 1 2

output

aedcbf

input

vwxyz 2 2 2

output

vwxyz

input

abcdef 3 1 2 3

output

fbdcea

规律：m次翻转顺序无关紧要，只可能出现自己或对称点的字符，

暴力统计翻转次数即可

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n;
char s[MAXN];
int m,a[MAXN];
bool b[MAXN]={0}; 
int main()
{
//  freopen("String.in","r",stdin);
  
  scanf("%s",s+1);
  n=strlen(s+1);
  scanf("%d",&m);
  For(i,m) scanf("%d",&a[i]);
  
  For(i,m)
  {
    b[a[i]]^=1;
  }
  
  bool flag=0;
  For(i,n/2)
  {
    flag^=b[i];
    if (flag) swap(s[i],s[n-i+1]); 
  }
  
  printf("%s\n",s+1);
    
  return 0;
}