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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.
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.
In the first line of the output print what Pasha‘s string s will look like after m days.
abcdef 1 2
aedcbf
vwxyz 2 2 2
vwxyz
abcdef 3 1 2 3
fbdcea
#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;
}
Codeforces Round #297 (Div. 2) B - Pasha and String
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原文地址：http://blog.csdn.net/u014303647/article/details/44752725
