CF1091G
题意
做法
令\(f(a)\)为输出\(sqrt~a\)得到的结果,也就是\(f(a)=x,.s.t~x^2\equiv a(mod~n)\)
随机选择\(x\in[1,n)\),得到\(f(x^2)\),令\(y=f(x^2)\)
令\(n=p_1p_2...p_k\),则\(y^2\equiv x^2(mod~p_1),y^2\equiv x^2(mod~p_2)...y^2\equiv x^2(mod~p_k)\)
对于\(y^2\equiv x^2(mod~p_i),y\equiv x~or~-x(mod~p_i)\)
\(\Pr \Big[ f(x^2) \equiv x \pmod{p_i} \Big| p_i \not| x \Big] = \frac 1 2\),\(\Pr \Big[ f(x^2) \equiv x \pmod n \Big| \gcd(x,n) = 1 \Big] = \frac 1 {2^k}\)
也就是若\((x,n)=1\),得到的\(y\),\(y\neq x\)的概率为\(1-\frac{1}{2^k}\),则\(x^2\equiv y^2(mod~n)\)
因为\(n=p_1p_2...p_k\),则必定有\(n=n_1n_2,n_1>1,n_2>1\),使得\(x\equiv y(mod~n_1),x\not\equiv y(mod~n_2),(n_1,n_2)=1\),
\(n_1|x-y,(n_1,n_2)=1\Longrightarrow (n,x-y)=n_1\)
题外话
一直忽略了不同的质因子...然后一直做不出...