2019ICPC南京网络赛C. Tsy's number 5

发表于 | 阅读数

把$i,j$分开

$\phi(i)$显然可以计数。$f[i]=\sum[\phi(k)=i]$

$h(x)= f(x)x2^{-x^2},g(x)=(\sqrt{2})^{(x)^2}$

其实这步已经可以卷了。进一步化简$i+j=t$

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 4e5 + 10;

const int mod = 998244353;
const int e2 = 116195171;
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

int inc(int x, int y) { return (x + y >= mod) ? (x + y - mod) : (x + y); }
int del(int x, int y) { return (x - y < 0) ? (x - y + mod) : (x - y); }
const int P = 998244353;
const int gi = 3;
int qpow(int a, ll x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mo;
        a = 1ll * a * a % mo;
        x >>= 1;
    }
    return res;
}

int rev[N];
void NTT(int *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        int tt = qpow(gi, (P - 1) / (l << 1), P);
        int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
        for (int i = 0; i < n; i += (l << 1))
        {
            int omega = 1;
            for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P)
            {
                int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
                A[i + j] = inc(x, y);
                A[i + j + l] = del(x, y);
            }
        }
    }
    int invv = qpow(n, P - 2, P);
    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i] = 1ll * A[i] * invv % P;
}
int initNTT(int n, int m)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    return ML;
}
void NTTX(int *a, int n, int *b, int m)
{
    int ML = initNTT(n, m);
    NTT(a, ML, 1);
    NTT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        a[i] = 1ll * a[i] * b[i] % P;
    NTT(a, ML, -1);
}
int pcnt, pr[N], npr[N], phi[N];

void Prime_init(int X)
{
    npr[1] = 1;
    phi[1] = 1;
    for (int i = 2; i <= X; i++)
    {
        if (!npr[i])
            pr[++pcnt] = i, phi[i] = i - 1;
        for (int j = 1; j <= pcnt && pr[j] * i <= X; j++)
        {
            npr[pr[j] * i] = 1;

            if (i % pr[j] == 0)
            {
                phi[i * pr[j]] = phi[i] * pr[j];
                break;
            }
            else
            {
                phi[i * pr[j]] = phi[i] * phi[pr[j]];
            }
        }
    }
}
int n, k;
int f[N], g[N], a[N], b[N], e2p[N];
int main()
{
    int T = read();
    Prime_init(1e5 + 10);
    while (T--)
    {
        memset(b, 0, sizeof(b));
        memset(f, 0, sizeof(f));
        memset(g, 0, sizeof(g));
        memset(a, 0, sizeof(a));
        int n = read();

        for (int i = 1; i <= n; i++)
            b[phi[i]]++;
        for (int i = 0; i <= n; i++)
            e2p[i] = qpow(e2, 1ll * i * i, mod);
        for (int i = 0; i <= 2 * n; i++)
            f[i] = qpow(e2, 1ll * i * i, mod);
        for (int i = 1; i <= n; i++)
        {
            g[i] = a[i] = 1ll * b[i] * i % mod * qpow(e2p[i], mod - 2, mod) % mod;
        }
        // reverse(g, g + 1 + n);
        NTTX(g, n + 1, a, n + 1);
        //reverse(g, g + 1 + n);
        // int duo = 0;
        // for (int i = 1; i <= n; i++)
        //     duo = inc(duo, 1ll * i * i % mod * b[i] % mod * b[i] % mod * qpow(2, 1ll * i * i, mod) % mod);

        int res = 0;
        for (int i = 1; i <= 2 * n; i++)
        {

            res = inc(res, 1ll * f[i] * g[i] % mod);
        }

        printf("%d\n", res);
    }
}