【NOIP2016练习】T1 string （计数）

题意：

思路：

 1 const mo=1000000007;
 2 var pow,f,exf:array[-100001..100001]of int64;
 3     n,k,i:longint;
 4     ans,x,y:int64;
 5     s,t:ansistring;
 6 
 7 function c(x,y:longint):int64;
 8 begin
 9  exit(f[x]*exf[y] mod mo*exf[x-y] mod mo);
10 end;
11 
12 function min(x,y:int64):int64;
13 begin
14  if x<y then exit(x);
15  exit(y);
16 end;
17 
18 begin
19  assign(input,'1.in'); reset(input);
20  assign(output,'1.out'); rewrite(output);
21  readln(n,k);
22  readln(s);
23  readln(t);
24  pow[0]:=1;
25  for i:=1 to n do pow[i]:=pow[i-1]*25 mod mo;
26  f[0]:=1;
27  for i:=1 to n do f[i]:=f[i-1]*i mod mo;
28  exf[0]:=1; exf[1]:=1;
29  for i:=2 to n do exf[i]:=exf[mo mod i]*(mo-mo div i) mod mo;
30  for i:=1 to n do exf[i]:=exf[i-1]*exf[i] mod mo;
31  ans:=1;
32  for i:=1 to n do
33  begin
34   x:=ord(t[i])-ord('a'); y:=ord(s[i])-ord('a');
35   ans:=(ans+min(x,y)*c(n-i,k-1) mod mo*pow[k-1] mod mo) mod mo;
36   if s[i]<t[i] then
37   begin
38    ans:=(ans+c(n-i,k)*pow[k] mod mo) mod mo;
39    ans:=(ans+(x-y-1)*c(n-i,k-1) mod mo*pow[k-1] mod mo) mod mo;
40   end;
41   if s[i]<>t[i] then dec(k);
42  end;
43  writeln(ans);
44  close(input);
45  close(output);
46 end.

 

posted on 2016-10-31 20:53  myx12345  阅读(175)  评论(0)  编辑  收藏  举报

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