BZOJ3288 Mato矩阵
题意
N×N方阵A,a(i,j)=gcd(i,j),求det(A)
N!>1000000
Tag 行列式,高斯消元,找规律,欧拉筛
这道题的分数-时间曲线十分畸形,除非你不会行列式和线性筛,否则就是想通100分否则0分(我们暂且认为高斯消元求行列式0分)
如果你打出前几个数不能发现什么规律,除了答案合成度非常高,要看出来就真是神牛了Orzzzz.
但是我们考虑基础行列式的变换,从第一行开始每行向下消元消成上三角矩阵.
我们发现a(i,i)=phi(i)之后得到答案Ans=Pi(phi(i)),欧拉-JZP大法好.
Mato三题还剩最后一题
program bzoj3288;
const maxn=1000017;
mo=1000000007;
var p:array[0..maxn] of boolean;
prime,phi:array[0..maxn] of longint;
n,cp:longint;
procedure sieve(n:longint);
var i,j:longint;
begin
fillchar(p,sizeof(p),true); p[1]:=false; cp:=0; phi[1]:=1;
for i:=1 to n do begin
if p[i] then begin
inc(cp); prime[cp]:=i;
phi[i]:=i-1;
end; for j:=1 to cp do if i*prime[j]<=n then begin
p[i*prime[j]]:=false; if i mod prime[j]=0 then begin
phi[i*prime[j]]:=phi[i]*prime[j]; break;
end else phi[i*prime[j]]:=phi[i]*phi[prime[j]];
end else break;
end;
end;
function piphi(n:longint):longint;
var i:longint;
begin
piphi:=1; for i:=1 to n do piphi:=int64(piphi)*phi[i]mod mo;
end;
begin
readln(n);
sieve(n);
writeln(piphi(n));
end.
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