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
初看是模拟题，结果看了一下时间复杂度，模拟坑定超时。
ai和s+1-ai是对称的，所以反转偶数次的相当于没翻转，奇数次的要翻转，所以我们只要统计每个字母翻转的次数，用前缀和处理。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
char s[maxn];
int m;
int sum[maxn];
int p[maxn];
int main()
{
int n;
int ss;
//freopen("in.txt","r",stdin);
while(~scanf("%s",s)){
scanf("%d",&m);
ss=strlen(s)-1;
memset(p,0,sizeof p);
for(int i=0;i<m;i++){
scanf("%d",&n);
p[n-1]++;
p[ss-n+1+1]--;
}
sum[0]=p[0];
if(sum[0]%2==1)swap(s[0],s[ss]);
for(int i=1;i<=ss/2;i++){
sum[i]=sum[i-1]+p[i];
if(sum[i]%2==1){
swap(s[i],s[ss-i]);
}
}
printf("%s\n",s);
}
}