[做题记录]数学#1

#include<bits/stdc++.h>
#define ll long long
#define N 22
#define P 998244353

ll t,p,q,r,l;

struct Po{
	ll cntu,cntr,sumi,sums,sqrs,prod;
	Po(){cntu = cntr = sumi = sums = sqrs = prod = 0;}
	Po operator + (Po b) {
		Po c;
		c.cntu=(cntu+b.cntu)%P,c.cntr=(cntr+b.cntr)%P;
		c.sumi=(sumi+b.sumi+cntr*b.cntr)%P;
		c.sums=(sums+b.sums+cntu*b.cntr)%P;
		c.sqrs=(sqrs+b.sqrs+((cntu*cntu)%P)*b.cntr+(2*cntu*b.sums)%P)%P;
		c.prod=((prod+b.prod+((cntu*cntr)%P)*b.cntr)%P+cntu*b.sumi+cntr*b.sums)%P;
		return c;
	}
}nu,nr,ans;

inline Po pow(Po a,ll k){
	Po res;
	while(k){
		if(k & 1){res = res + a;}
		a = a + a;
		k >>= 1;
	}
	return res;
}

inline ll div(ll a,ll b,ll c,ll d){return ((long double)1.0 * a * b + c) / d;}

inline Po solve(ll p,ll q,ll r,ll l,Po a,Po b){
	if(!l)return Po();
	if(p >= q)return solve(p % q,q,r,l,a,pow(a,p / q) + b);
	ll m = div(l,p,r,q);
	if(!m)return pow(b,l);
	ll cnt = l - div(q,m,-r - 1,p);
	return pow(b,(q - r - 1) / p) + a + solve(q,p,(q - r - 1) % p,m - 1,b,a) + pow(b,cnt);
}

int main(){
	scanf("%lld",&t);
	while(t -- ){
		scanf("%lld%lld%lld%lld",&l,&p,&r,&q);
		nu.cntu = 1,nu.cntr = nu.sumi = nu.sums = nu.sqrs = nu.prod = 0;
		nr.cntu = nr.sums = nr.sqrs = nr.prod = 0,nr.sumi = nr.cntr = 1;
		ans = pow(nu,r / q) + solve(p,q,r % q,l,nu,nr);
		printf("%lld %lld %lld\n",(ans.sums+r/q)%P,(ans.sqrs+((r/q)%P)*((r/q)%P))%P,ans.prod);		
	}
}

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int t;
long long max_factor, n;

long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}

long long quick_pow(long long x, long long p, long long mod) {  //快速幂
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (__int128)ans * x % mod;
    x = (__int128)x * x % mod;
    p >>= 1;
  }
  return ans;
}

bool Miller_Rabin(long long p) {  //判断素数
  if (p < 2) return 0;
  if (p == 2) return 1;
  if (p == 3) return 1;
  long long d = p - 1, r = 0;
  while (!(d & 1)) ++r, d >>= 1;  //将d处理为奇数
  for (long long k = 0; k < 10; ++k) {
    long long a = rand() % (p - 2) + 2;
    long long x = quick_pow(a, d, p);
    if (x == 1 || x == p - 1) continue;
    for (int i = 0; i < r - 1; ++i) {
      x = (__int128)x * x % p;
      if (x == p - 1) break;
    }
    if (x != p - 1) return 0;
  }
  return 1;
}

long long Pollard_Rho(long long x) {
  long long s = 0, t = 0;
  long long c = (long long)rand() % (x - 1) + 1;
  int step = 0, goal = 1;
  long long val = 1;
  for (goal = 1;; goal *= 2, s = t, val = 1) {  //倍增优化
    for (step = 1; step <= goal; ++step) {
      t = ((__int128)t * t + c) % x;
      val = (__int128)val * abs(t - s) % x;
      if ((step % 127) == 0) {
        long long d = gcd(val, x);
        if (d > 1) return d;
      }
    }
    long long d = gcd(val, x);
    if (d > 1) return d;
  }
}

void fac(long long x) {
  if (x <= max_factor || x < 2) return;
  if (Miller_Rabin(x)) {              //如果x为质数
    max_factor = max(max_factor, x);  //更新答案
    return;
  }
  long long p = x;
  while (p >= x) p = Pollard_Rho(x);  //使用该算法
  while ((x % p) == 0) x /= p;
  fac(x), fac(p);  //继续向下分解x和p
}

int main() {
  scanf("%d", &t);
  while (t--) {
    srand((unsigned)time(NULL));
    max_factor = 0;
    scanf("%lld", &n);
    fac(n);
    if (max_factor == n)  //最大的质因数即自己
      printf("Prime\n");
    else
      printf("%lld\n", max_factor);
  }
  return 0;
}