tyvj1068 STR

给你两个串A，B，可以得到从A的任意位开始的子串和B匹配的长度。
给定K个询问，对于每个询问给定一个x，求出匹配长度恰为x的位置有多少个。
N，M，K<=200000

字符串的处理，用KMP喽，首先求出模式串B的next[]，然后进行一遍AB匹配，记录a[i]匹配的最大长度，那么如果a[i]可以匹配l[i]的长度，answer[i]表示匹配长度为i的串的数量，则answer[next[i]]的匹配长度也是满足answer[i]的，所以只要倒着循环一遍，把answer[next[i]]加到answer[i]中即可。这时的answer[i]包含了长度为i+1的情况，所以要输出answer[k]-answer[k+1].

program tyvj1068(input,output);
var
   a,b      : ansistring;
   next,l : array[0..201000] of longint;
   answer : array[0..201000] of longint;
   i,j,x  : longint;
   n,m,k  : longint;
   ch      : char;
begin
   readln(n,m,k);
   a:='';
   b:='';
   for i:=1 to n do
   begin
      read(ch);
      a:=a+ch;
   end;
   readln;
   for i:=1 to m do
   begin
      read(ch);
      b:=b+ch;
   end;
   readln;
   fillchar(next,sizeof(next),0);
   next[1]:=0;
   j:=0;
   for i:=2 to m do
   begin
      while (j>0)and(b[j+1]<>b[i]) do
     j:=next[j];
      if b[j+1]=b[i] then
     inc(j);
      next[i]:=j;
   end;
   j:=0;
   for i:=1 to n do
   begin
      while (j>0)and(b[j+1]<>a[i]) do
     j:=next[j];
      if b[j+1]=a[i] then
     inc(j);
      l[i]:=j;
      if j=m then
     j:=next[j];
   end;
   fillchar(answer,sizeof(answer),0);
   for i:=1 to n do
      inc(answer[l[i]]);
   for i:=n downto 1 do
      inc(answer[next[i]],answer[i]);
   for i:=1 to k do
   begin
      readln(x);
      writeln(answer[x]-answer[x+1]);
   end;
end.