Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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 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 Input
abcdef 1 2
aedcbf
vwxyz 2 2 2
vwxyz
abcdef 3 1 2 3
fbdcea
题解：
对于一个字符串，给m个询问，每次交换从i到n-i之间的字符，由于大的区间包含小的区间，只需要求交换的总次数就好了，奇数交换对称的两个数，偶数不交换；
代码：
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
using namespace std;
const int MAXN = 2e5 + ;
int tree[MAXN];
char s[MAXN];
int main(){
int m, x;
while(~scanf("%s", s + )){
scanf("%d", &m);
memset(tree, , sizeof(tree));
while(m--){
scanf("%d", &x);
tree[x]++;
}
int len = strlen(s + );
for(int i = ; i <= len / ; i++){
tree[i] += tree[i - ];
}
for(int i = ; i <= len / ; i++){
if(tree[i] & )swap(s[i], s[len - i + ]);
}
printf("%s\n",s + );
}
return ;
}