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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 + 100; int tree[MAXN]; char s[MAXN]; int main(){ int m, x; while(~scanf("%s", s + 1)){ scanf("%d", &m); memset(tree, 0, sizeof(tree)); while(m--){ scanf("%d", &x); tree[x]++; } int len = strlen(s + 1); for(int i = 2; i <= len / 2; i++){ tree[i] += tree[i - 1]; } for(int i = 1; i <= len / 2; i++){ if(tree[i] & 1)swap(s[i], s[len - i + 1]); } printf("%s\n",s + 1); } return 0; }
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原文地址：http://www.cnblogs.com/handsomecui/p/5414774.html
