2017.6.4 入门组 NO.5——序列

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f[i,j*k]:=f[i,j*k]+f[i-1,j]； 
设f[i,j]表示前i个，当那一位数字为j*k的时候的最大好序列个数

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代码如下：

const
        maxn=2000;
        p=1000000007;
var
        f:array [1..maxn,1..maxn] of longint;
        i,j,k,n,m:longint;
        ans:int64;
begin
        readln(n,m);
        for i:=1 to n do
         f[1,i]:=1;
        for i:=2 to m do
         for j:=1 to n do
          for k:=1 to n div j do
            f[i,j*k]:=(f[i,j*k]+f[i-1,j]) mod p;
        for i:=1 to n do ans:=(ans+f[m,i]) mod p;
        writeln(ans);
end.