其实真的不难…自己找一下规律吧.因为有mod m嘛…然后就几个一循环而已…
{ SGU 181; X-Sequence - sqybi’s code}//for my winstyprogram sgu181_sqybi; const mm = 1000;
var n, i: longint; x, a, b, c, m, stop, r: int64; f, t: array[0..mm]of longint;
begin readln(x, a, b, c, m, n); if n = 0 then begin writeln(x); halt; end; fillchar(f, sizeof(f), 0); stop := 0; […]
bt的数学题.题解看这里:http://www.mydrs.org/dvp/dispbbs.php?boardid=13&id=436842楼说的还算比较详细…我刚开始把题目里P的定义的乘法看成了加法所以一直不懂2楼说的,后来看题之后才发现自己看错了…
{ SGU 169; Numbers - sqybi’s code - Maths}//for my winstyprogram sgu169_sqybi; var n, t: longint;
begin readln(n); if n = 1 then writeln(8) else begin t := 1; if (n - 1) mod 3 = 0 then t := t + 2; if (n - 1) mod 6 = 0 then t := […]
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(82 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 […]
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