蒟蒻的第一篇博文

有点小激动，因为百度空间的革新，近期无法记录一些好的解题思路了，这才想起来有csdn账号，好久不用了...BZOJ1005[HNOI2008]明明的烦恼 一道SB题，教会了我们一个神奇的序列--purfer序列。so，这道题变成统计序列的个数，且每个点的度数减一即为每个点在purfer序列中出现的次数。最后组合数搞一搞就好...

purfer序列详解

/**************************************************************    Problem: 1005    User: liuxin    Language: Pascal    Result: Accepted    Time:236 ms    Memory:304 kb****************************************************************/const    size=168;    data:array[1..size]of longint=(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,  79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,  193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,  313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,  443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,  587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,  719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857 ,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997);var i,j,k,len,sum,n,a:longint;    ans,t:array[0..10001]of longint;    procedure C(n,m:longint);    var i,j,p:longint;    begin        for i:=m+1 to n do begin            p:=i; j:=1;            while p>1 do begin                while p mod data[j]=0 do begin                    inc(ans[j]);                    p:=p div data[j];                end;                inc(j);            end;        end;        for i:=1 to n-m do begin            p:=i; j:=1;            while p>1 do begin                while p mod data[j]=0 do begin                    dec(ans[j]);                    p:=p div data[j];                end;                inc(j);            end;        end;    end;begin    readln(n); len:=n-2; sum:=0;    for i:=1 to n do begin        readln(a);        if (a=0)or(len-a+1<0) then begin            writeln(0);            halt;        end;        if a=-1 then begin            inc(sum);            continue;        end;        dec(a);        C(len,a); len:=len-a;    end;    if len>0 then begin        j:=1;        while sum>1 do begin            while sum mod data[j]=0 do begin                ans[j]:=ans[j]+len;                sum:=sum div data[j];            end;            inc(j);        end;    end;    t[1]:=1; k:=1;    for i:=1 to size do begin        while ans[i]>0 do begin            dec(ans[i]);            for j:=1 to k do t[j]:=t[j]*data[i];            for j:=1 to k do            if t[j]>9 then begin                t[j+1]:=t[j+1]+t[j] div 10;                t[j]:=t[j] mod 10;            end;            while t[k+1]>0 do begin                inc(k);                t[k+1]:=t[k+1]+t[k] div 10;                t[k]:=t[k] mod 10;            end;        end;    end;    for i:=k downto 1 do write(t[i]); writeln;end.