QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#570381#9309. GraphtarjenWA 44ms56708kbC++202.9kb2024-09-17 15:33:262024-09-17 15:33:29

Judging History

你现在查看的是最新测评结果

  • [2024-09-17 15:33:29]
  • 评测
  • 测评结果:WA
  • 用时:44ms
  • 内存:56708kb
  • [2024-09-17 15:33:26]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
//??????? n^1/3 ?????????n^2/3???????????
const int N  = 1e7;
const int M  = 2;         //?????????????
const int PM = 2 * 3 * 5; //???????????????????17
ll n;
bool np[N];
int prime[N], pi[N];
int phi[PM + 1][M + 1], sz[M + 1];

int getprime() {
    int cnt = 0;
    np[0] = np[1] = true;
    pi[0] = pi[1] = 0;
    for (int i = 2; i < N; ++i) {
        if (!np[i]) prime[++cnt] = i;
        pi[i] = cnt;
        for (int j = 1; j <= cnt && i * prime[j] < N; ++j) {
            np[i * prime[j]] = true;
            if (i % prime[j] == 0) break;
        }
    }
    return cnt;
}

void init() {
    getprime();
    sz[0] = 1;
    for (int i = 0; i <= PM; ++i) phi[i][0] = i;
    for (int i = 1; i <= M; ++i) {
        sz[i] = prime[i] * sz[i - 1];
        for (int j = 1; j <= PM; ++j) phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];
    }
}

int sqrt2(ll x) {
    ll r = (ll)sqrt(x - 0.1);
    while (r * r <= x) ++r;
    return int(r - 1);
}

int sqrt3(ll x) {
    ll r = (ll)cbrt(x - 0.1);
    while (r * r * r <= x) ++r;
    return int(r - 1);
}

ll getphi(ll x, int s) {
    if (s == 0) return x;
    if (s <= M) return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];
    if (x <= prime[s] * prime[s]) return pi[x] - s + 1;
    if (x <= prime[s] * prime[s] * prime[s] && x < N) {
        int s2x = pi[sqrt2(x)];
        ll ans  = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;
        for (int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]];
        return ans;
    }
    return getphi(x, s - 1) - getphi(x / prime[s], s - 1);
}

ll getpi(ll x) {
    if (x < N) return pi[x];
    ll ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;
    for (int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + 1;
    return ans;
}

ll lehmer_pi(ll x) { //????n???????
    if (x < N) return pi[x];
    int a  = (int)lehmer_pi(sqrt2(sqrt2(x)));
    int b  = (int)lehmer_pi(sqrt2(x));
    int c  = (int)lehmer_pi(sqrt3(x));
    ll sum = getphi(x, a) + (ll)(b + a - 2) * (b - a + 1) / 2;
    for (int i = a + 1; i <= b; i++) {
        ll w = x / prime[i];
        sum -= lehmer_pi(w);
        if (i > c) continue;
        ll lim = lehmer_pi(sqrt2(w));
        for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1);
    }
    return sum;
}
const ll mod =998244353;
ll ksm(ll x,ll k){
    ll res=1;
    while(k){
        if(k&1)res=res*x%mod;
        x=x*x%mod;
        k/=2;
    }
    return res;
}
int main()
{
    init();
    ll n;cin>>n;
    ll ans=1;
    auto solve = [&](ll n){
        ll res=1;
        if(n==1)return res;
        ll p=lehmer_pi(n)-lehmer_pi(n/2)+1;
        assert(p>=2);
        if(n-p>0)res=ksm(n,p+1-2);
        else res=ksm(n,p-2);
        if(n-p>0)res=res*(n-p)%mod;
        return res;
    };
    for(ll l=1,r;l<=n;l=r+1){
        r=min(n/(n/l),n);
        ll x=n/l;
        ans=ans*ksm(solve(x),r-l+1);
    }
    cout<<ans;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 35ms
memory: 56668kb

input:

4

output:

8

result:

ok answer is '8'

Test #2:

score: 0
Accepted
time: 44ms
memory: 56708kb

input:

2

output:

1

result:

ok answer is '1'

Test #3:

score: -100
Wrong Answer
time: 34ms
memory: 56708kb

input:

123

output:

0

result:

wrong answer expected '671840470', found '0'