THUPC2017 小 L 的计算题

求 $k=1,2,cdots,n space space sumlimits_{i=1}^n a_i^k$

$n leq 2 imes 10^5$

sol：

时隔多年终于卡过去了

之前 $O(nlog^2n) + O(nlogn)$ 卡了我的 $O(nlog^2n) + O(nlog^2n)$ ，有点自闭

然后 fread + 编译优化 + 预处理单位根 + 不在 fft 里计算 rev 数组大力卡进时限

#include <bits/stdc++.h>
#define LL long long
#define rep(i, s, t) for (register int i = (s), i##end = (t); i <= i##end; ++i)
#define dwn(i, s, t) for (register int i = (s), i##end = (t); i >= i##end; --i)
using namespace std;
const int Size=1<<16;
char buffer[Size],*head,*tail;
inline char Getchar() {
    if(head==tail) {
        int l=fread(buffer,1,Size,stdin);
        tail=(head=buffer)+l;
    }
    if(head==tail) return -1;
    return *head++;
}
inline int read() {
    int x=0,f=1;char c=Getchar();
    for(;!isdigit(c);c=Getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=Getchar()) x=x*10+c-'0';
    return x*f;
}
const int mod = 998244353, maxn = 800010;
int a[maxn], r[maxn], lg[maxn], n, k;
inline int skr(int x, int t) {
    int res = 1;
    while (t) {
        if (t & 1)
            res = 1LL * res * x % mod;
        x = 1LL * x * x % mod;
        t = t >> 1;
    }
    return res;
}
int wn[maxn], iwn[maxn];
void init(int n) {
    wn[0] = iwn[0] = 1; 
    rep(i, 1, n-1) wn[i] = skr(3, (mod - 1) / (i << 1));
    rep(i, 1, n-1) iwn[i] = skr(332748118, (mod - 1) / (i << 1));
}
inline void fft_init(int n) {rep(i, 0, n - 1) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (lg[n] - 1));}
inline void fft(int *a, int n, int type) {
    rep(i, 0, n - 1) if (i < r[i]) swap(a[i], a[r[i]]);
    for (int i = 1; i < n; i <<= 1) {
        //int wn = skr(3, (mod - 1) / (i << 1));
        //if (type == -1)
        //    wn = skr(wn, mod - 2);
        int twn = (type == -1) ? iwn[i] : wn[i];
        for (int j = 0; j < n; j += (i << 1)) {
            int w = 1;
            for (int k = 0; k < i; k++, w = 1LL * w * twn % mod) {
                int x = a[j + k], y = 1LL * w * a[j + k + i] % mod;
                a[j + k] = (x + y) % mod;
                a[j + k + i] = (x - y + mod) % mod;
            }
        }
    }
    if (type == -1) {
        int inv_n = skr(n, mod - 2);
        rep(i, 0, n - 1) a[i] = 1LL * a[i] * inv_n % mod;
    }
}
int A[maxn], B[maxn];
int C[maxn], D[maxn];
int mul(int *A, int *B, int len) {
    fft_init(len);
    // fft_init(len);
    fft(A, len, 1);
    // for(int i=0;i<len;i++)cout<<A[i]<<" ";
    // cout<<endl;
    fft(B, len, 1);
    for (int i = 0; i < len; i++) A[i] = (LL)A[i] * B[i] % mod;
    fft(A, len, -1);
    --len;
    while (!A[len]) --len;
    return len;
}
vector<int> poly[maxn];
int solve(int l, int r) {
    if (l == r)
        return poly[l].size() - 1;
    int mid = (l + r) >> 1;
    int ls = solve(l, mid), rs = solve(mid + 1, r);
    int L = 1;
    for (; L <= ls + rs; L <<= 1)
        ;

    for (int i = 0; i <= ls; i++) A[i] = poly[l][i];
    for (int i = ls + 1; i < L; i++) A[i] = 0;

    for (int i = 0; i <= rs; i++) B[i] = poly[mid + 1][i];
    for (int i = rs + 1; i < L; i++) B[i] = 0;
    poly[l].clear();
    poly[mid + 1].clear();

    L = mul(A, B, L);
    for (int i = 0; i <= L; i++) poly[l].push_back(A[i]);
    return L;
}
int g[maxn], f[maxn];
void mulfac(int *A, int *B, int len) {
    fft_init(len);
    fft(A, len, 1);
    fft(B, len, 1);
    for (int i = 0; i < len; i++) A[i] = 1LL * A[i] * B[i] % mod;
    fft(A, len, -1);
}
void cdq_fft(int *f, int *g, int l, int r) {
    if (l == r) {
        (f[l] += (1LL * l * g[l] % mod)) %= mod;
        return;
    }
    int mid = (l + r) >> 1;
    cdq_fft(f, g, l, mid);
    int len = 1, ls = 0, rs = 0;
    // for(;len <= ((r - l + mid)<<1);len <<= 1);
    // for(int i=0;i<len;i++)A[i] = B[i] = 0;
    for (int i = l; i <= mid; i++) C[ls++] = f[i];
    for (int i = 1; i <= r - l; i++) D[rs++] = g[i];
    for (; len <= (ls + rs - 1); len <<= 1)
        ;
    mulfac(C, D, len);
    for (int i = mid + 1; i <= r; i++) f[i] = (f[i] + C[i - l - 1]) % mod;
    for (int i = 0; i < len; i++) C[i] = D[i] = 0;
    cdq_fft(f, g, mid + 1, r);
}
int main() {
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    lg[0] = -1; init(1 << 19);
    rep(i, 1, maxn - 1) lg[i] = lg[i >> 1] + 1;
    int T = read();
    while (T--) {
        int ans = 0;
        n = read();
        for (int i = 1; i <= n; i++) {
            a[i] = read();
            if(a[i] >= mod) a[i] -= mod;
            poly[i].push_back(1);
            poly[i].push_back(a[i]);
        }
        solve(1, n);
        for (int i = 0; i < poly[1].size(); i++) g[i] = (((i & 1) ? 1 : (-1)) * poly[1][i] + mod) % mod;
        poly[1].clear();
        cdq_fft(f, g, 0, n);
        for (int i = 1; i <= n; i++) ans ^= f[i];
        memset(f, 0, sizeof(f));
        memset(g, 0, sizeof(g));
        memset(A, 0, sizeof(A));
        memset(B, 0, sizeof(B));
        memset(C, 0, sizeof(C));
        memset(D, 0, sizeof(D));
        cout << ans << endl;
    }
}

 然而这种 shabi 题为什么我能写 6K，给镘写也就 100 行，我菜的真实