cf- 297 < b > -- 区间翻转操作的优化

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 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

 *****************************************************

题目简述：

给一串字符，给m步操作，每步操作是将从字符串的第n个开始到第 lenth-n+1  的区间进行翻转；求最后的字符串；

*******************************************************

本题暴力，用一个数组记录该字符背反转的次数，被翻转偶数次就相当与没有翻转，基数此就相当于和对位的字符进行一次交换；

刚开始我是 n^2 的算法，从第n个到第 lenth - n -1 的区间扫一遍加一，这样梅雨哦必要，因为后面一个翻转的次数总是大于等于前面一个的翻转次数，所以只由前一个累加过来就得到翻转次数；

首先扫一遍操作数组（先排个序），将区间的首尾位置标记出来，（就是加一），区间首之间的元素翻转次数是和前一个区间首次数相同的，然后来扫一遍数组，将各个元素的翻转次数累加出来，前后是堆成的，所以是要扫一半就可以了，同时处理前后两个对应的节点；

最后输出的时候检验此数，偶数鸳鸯输出，奇数输出对位元素；

代码如下：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<string>
#include<vector>
using namespace std;
char s[200010];
int m,sl=0,a[200010]={0},d[100010]={0};
int main()
{
    memset(a,0,sizeof(a));
    scanf("%s",s);
    scanf("%d",&m);
    sl=strlen(s);
    for(int i=0;i<m;i++)
    {
        scanf("%d",&d[i]);
    }
    sort(d,d+m);
    for(int i=0;i<m;i++)
    {
        a[d[i]-1]++;
        a[sl-d[i]]++;
    }
    for(int i=1;i<sl/2;i++)
    {
        a[i]+=a[i-1];
        a[sl-i-1]+=a[sl-i];
    }
    for(int i=0;i<sl;i++)
    {
        if(a[i]%2==0)
            printf("%c",s[i]);
        else
            printf("%c",s[sl-i-1]);
    }
    printf("
");
    return 0;
}