字符串处理 Codeforces Round #297 (Div. 2) B. Pasha and String

题目传送门

 /*
     题意：给出m个位置，每次把[p,len-p+1]内的字符子串反转，输出最后的结果
     字符串处理：朴素的方法超时，想到结果要么是反转要么没有反转，所以记录
                 每个转换的次数，把每次要反转的反转就不超时了:)
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 using namespace std;

 const int MAXN = 2e5 + ;
 const int INF = 0x3f3f3f3f;
 char s[MAXN];
 int to[MAXN];

 int main(void)        //Codeforces Round #297 (Div. 2) B. Pasha and String
 {
     int m;
     scanf ("%s", s + );
     int len = strlen (s + );
     scanf ("%d", &m);
     while (m--)
     {
         int p;    scanf ("%d", &p);
         int q = len - p + ;
         to[p]++;
     }

     int sum = ;
     for (int i=; i<=len/; ++i)
     {
         sum += to[i];
         if (sum & )    swap (s[i], s[len-i+]);
     }
     printf ("%s", s + );

     return ;
 }

 /*
 abcdef
 1
 2
 vwxyz
 2
 2 2
 abcdef
 3
 1 2 3
 */