稍微推导一下就可以得到一个递推式(这里直接借用WindyWinter的式子):pos[1,1]=1pos[n,q]=n-k+pos[k,k-q+1] (q<=k) =pos[n-k,n-q+1] (q>k)
求pos[n, q]即可.
{ SGU 175; Encoding - sqybi’s code}//for my winstyprogram sgu175_sqybi; var n, t, q, k: longint;
begin readln(n, q); t := 0; while n <> 1 do begin k := n div 2; if q <= k then begin t := t + n - k; n := k; q […]
只说一句:图像法解决概率问题真好用.公式见程序.
{ SGU 144; Meeting - sqybi’s code - Maths}//for my winstyprogram sgu144_sqybi; var x, y, z: double;
begin readln(x, y, z); writeln((1-sqr((y-x)*60-z)/sqr((y-x)*60)):0:7); end.
阅读(122 次)
很简单的二分答案,不过要注意一点,题目所说的自然数不包括0(也就是说,如果输入是0,需要输出1.这里WA了2次).
{ SGU 154; Factorial - sqybi’s code - 二分答案}//for my winstyprogram sgu154_sqybi; var n, l, r, mid, i, t, ans: int64;
begin readln(n); ans := 0; l := 1; r := n * 5; while l <= r do begin mid := (l + r) div 2; i := 5; t := 0; while […]
n皇后问题,其中n<=50,而且棋盘上有一些部分已经放置皇后.Dancing Links算法搞掉,另外jl大牛似乎用一个改进的DLX算法(貌似是一个十八向的链表…)用极快的速度AC了这道题…另:C++比Pascal快一倍,真TMD无语…
{ SPOJ 1771; Yet Another N-Queen Problem - sqybi’s code - DLX}//for my winstyprogram nqueen_sqybi; const nn = 50; tt = nn * 6 - 2 + nn * nn * 4;
var n, i, j, k, temp, tot: longint; a: array[1..nn]of longint; col, row, u, d, l, r, size: array[0..tt]of longint; used: […]
水题.求出中位数即可.
根据WindyWinter的blog,有两点注意.
Pascal注意,读入n对数时,应该用read而不是readln,因为虽然样例数据有换行,但题目中的描述是没有换行的.否则会像我一样WA on 14.
C++注意,iostream会TLE on 12.
{ SGU 114; Telecasting station - sqybi’s code}//for my winstyprogram sgu114_sqybi; const nn = 15000;
type rec = record p, v: longint; end;
var s, t, n, i: longint; a: array[1..nn]of rec;
procedure qsort(l, r: longint); var i, j, d: longint; t: rec; begin if l >= r […]
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