题意转化一下就是寻找一个数P，要求P质因素分解完后，质因素没有重复，还要保证abs(P*P-x)最小。
暴力，在sqrt(x)附近向下向上分别枚举一下。
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<map> #include<set> #include<queue> #include<stack> #include<ctime> #include<iostream> using namespace std; typedef long long LL; const double pi=acos(-1.0),eps=1e-8; void File() { freopen("D:\in.txt","r",stdin); freopen("D:\out.txt","w",stdout); } inline int read() { char c = getchar(); while(!isdigit(c)) c = getchar(); int x = 0; while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); } return x; } const int S=20; LL mult_mod(LL a,LL b,LL c) { a%=c; b%=c; LL ret=0; while(b) { if(b&1){ret+=a;ret%=c;} a<<=1; if(a>=c)a%=c; b>>=1; } return ret; } LL pow_mod(LL x,LL n,LL mod) { if(n==1)return x%mod; x%=mod; LL tmp=x,ret=1; while(n) { if(n&1) ret=mult_mod(ret,tmp,mod); tmp=mult_mod(tmp,tmp,mod); n>>=1; } return ret; } bool check(LL a,LL n,LL x,LL t) { LL ret=pow_mod(a,x,n); LL last=ret; for(int i=1;i<=t;i++) { ret=mult_mod(ret,ret,n); if(ret==1&&last!=1&&last!=n-1) return true; last=ret; } if(ret!=1) return true; return false; } bool Miller_Rabin(LL n) { if(n<2)return false; if(n==2)return true; if((n&1)==0) return false; LL x=n-1,t=0; while((x&1)==0){x>>=1;t++;} for(int i=0;i<S;i++) { LL a=rand()%(n-1)+1; if(check(a,n,x,t)) return false; } return true; } LL factor[100]; int tol; LL gcd(LL a,LL b) { if(a==0)return 1; if(a<0) return gcd(-a,b); while(b) { LL t=a%b; a=b; b=t; } return a; } LL Pollard_rho(LL x,LL c) { LL i=1,k=2,x0=rand()%x,y=x0; while(1) { i++; x0=(mult_mod(x0,x0,x)+c)%x; LL d=gcd(y-x0,x); if(d!=1&&d!=x) return d; if(y==x0) return x; if(i==k){y=x0;k+=k;} } } void findfac(LL n) { if(Miller_Rabin(n)) { factor[tol++]=n; return; } LL p=n; while(p>=n)p=Pollard_rho(p,rand()%(n-1)+1); findfac(p); findfac(n/p); } int T; LL x,ans; int main() { srand(time(NULL)); scanf("%d",&T); while(T--) { scanf("%lld",&x); if(x==1) {printf("3 "); continue;} if(x==2) {printf("2 "); continue;} if(x==3) {printf("1 "); continue;} LL n=(LL)sqrt(1.0*x); ans=x; for(LL i=n;i>=1;i--) { tol=0; findfac(i); sort(factor,factor+tol); bool fail=0; for(int j=0;j<tol-1;j++) if(factor[j]==factor[j+1]) fail=1; if(fail==1) continue; else { ans=abs(x-i*i); break;} } for(LL i=n+1;;i++) { tol=0; findfac(i); sort(factor,factor+tol); bool fail=0; for(int j=0;j<tol-1;j++) if(factor[j]==factor[j+1]) fail=1; if(fail==1) continue; else { ans=min(ans,abs(x-i*i)); break;} } printf("%lld ",ans); } return 0; }