Codeforces round#297 Div2

B:

Test instructions: Given a string, and given m digits

For each of the digital AI meanings, flip the AI to the n-ai+1 string again.

A string that requires the output to be flipped after M-times.

The idea is to determine the position of the first character is reversed odd number of times, or even a few times on it, flipped an even number of times represents no flip

The idea is to store the m numbers down and then sort them. For each str[i], 1<=i<strlen (str),

Using Index=upper_bound (a,a+m,i), the return value of Upper_bound (A,a+m,i) is the first element iterator in the returned array that is larger than I, and if not, returns the iterator that points to the last element

So index-a represents the number of digits in array A that are smaller than I.

So if the index-a is odd, then the position I flips the odd number of times, the output STR[N-I+1]

If it is even, the code position I flips an even number of times, output STR[I]

1#include <stdio.h>2#include <string.h>3#include <algorithm>4 using namespacestd;5 Const intN =200000+Ten;6 CharStr[n];7 intA[n];8 BOOLVis[n];9 Ten intMain () One { A     intM,i,x,n; -scanf"%s", str+1); -scanf"%d",&m); then = strlen (str+1); -      for(i=1; i<=m; ++i) -scanf"%d",&a[i]); -Sort (A +1, a+m+1); +     intnn = n/2,*index; -      for(i=1; i<=nn; ++i) +     { A          atindex = Upper_bound (A +1, a+m+1, i); -         if((index-a-1)%2==1) -         { -printf"%c", str[n-i+1]); -vis[n-i+1] =true; -         } in         Else -printf"%c", Str[i]); to          +     } -      for(i=nn+1; i<=n; ++i) the     { *         if(Vis[i]) $printf"%c", str[n-i+1]);Panax Notoginseng         Else -printf"%c", Str[i]); the     } +Puts""); A}

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But watching someone else.

Codeforces round#297 Div2