2017.3.18 序列

分析：设f[i,j]为前i个数组成序列，最后一位为j的值，转移显然。

代码：

const
  p=1000000007;
var
 n,k,i,j,m,ans:longint;
 f:array [0..2000,0..2000] of longint;
begin
 read(n,m);
 for i:=1 to n do
  f[i,1]:=1;
  for j:=2 to m do
   for i:=1 to n do
    for k:=1 to n div i do
     begin
     if f[k*i,j-1]=0 then break;
     f[i,j]:=(f[i*k,j-1]+f[i,j]) mod p;
     end;
 for i:=1 to n do
  ans:=(ans+f[i,m]) mod p;
 writeln(ans);
end.