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
这题和涂气球那道类似,用了叠加的思想。
#include<stdio.h> #include<string.h> int d[200005]; char str[200005]; int main() { int n,m,i,j,a,len,s; char c; while(scanf("%s",str+1)!=EOF) { len=strlen(str+1); scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d",&a); d[a]++; } s=0; for(i=1;i<=len/2;i++) { s=s+d[i]; d[i]=s; } for(i=1;i<=len/2;i++) { if(d[i]%2==1) { c=str[i]; str[i]=str[len+1-i]; str[len+1-i]=c; } } //printf("%d %d %d",d[1],d[2],d[3]); for(i=1;i<=len;i++) { printf("%c",str[i]); } printf(" "); } }