i,n:longint;
a:array[1..1000] of longint;
begin
read(n);
a[1]:=1;a[2]:=2;
for i:=3 to n do
if i mod 2<>0 then a[i]:=a[i-1]
else a[i]:=a[i-1]+a[i div 2];
end;
write(a[n]);
end.
本文用了递推的思想!!!!!!!!!!!!!
真棒,比给
记忆化搜索
快速帮助 | 运行状态 | 反馈举报 | 关于我们 | 免责声明 | 浙ICP备11060257号 Processed in 0.0037 Second(s) Copyright (C) RQNOJ 2007-2019. All Rights Reserved.