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最新推荐文章于 2022-01-24 22:40:17 发布

转载最新推荐文章于 2022-01-24 22:40:17 发布 · 56 阅读

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原文链接：http://www.cnblogs.com/exponent/archive/2011/08/13/2137239.html

题意:求C(N,M)的约数个数.

分析:暴力分解质因数会TLE.所以先预处理.

记cnt[i,j]为i的阶乘中含第j个质数的个数(这样做要先筛素数).

然后C(N,M)=N!/(M!*(N-M)!)中含第i个质数的个数就是cnt[n,i]-cnt[m,i]-cnt[n-m,i].

这样就OK了(刷到了pascal的rank1).

code:

const MAX=440;
var   Prime:array[0..MAX] of longint;
      v:array[0..MAX] of boolean;
      cnt:array[0..max,0..100] of longint;
      size,n,m,i,j,num,tmp:longint;
      ans:qword;

      procedure GetPrime;
      begin
            size:=0;
            for i:=2 to MAX do
            begin
                  if not v[i] then
                  begin
                        inc(size);
                        prime[size]:=i;
                  end;
                  j:=1;
                  while (j<=size)and(prime[j]*i<MAX) do
                  begin
                        v[i*prime[j]]:=true;
                        if i mod prime[j]=0 then break;
                        inc(j);
                  end;
            end;
      end;

begin
      GetPrime;
      for i:=1 to size do
      begin
            cnt[1,i]:=0;
            for j:=2 to max do
            begin
                  tmp:=0;
                  num:=j;
                  while num mod prime[i]=0 do
                  begin
                        num:=num div prime[i];
                        inc(tmp);
                  end;
                  cnt[j,i]:=cnt[j-1,i]+tmp;
            end;
      end;

      while not seekeof do
      begin
            readln(n,m); 
            ans:=1;
            for i:=1 to size do
            begin
                  tmp:=cnt[n,i]-cnt[m,i]-cnt[n-m,i];
                  ans:=qword(ans)*qword(tmp+1);
            end;
            writeln(ans);
      end;
end.

转载于:https://www.cnblogs.com/exponent/archive/2011/08/13/2137239.html