THUPC2017 小 L 的计算题

求 $k=1,2,\cdots,n \space \space \sum\limits_{i=1}^n a_i^k$

$n \leq 2 \times 10^5$

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=<<;
char buffer[Size],*head,*tail;
inline char Getchar() {
    if(head==tail) {
        int l=fread(buffer,,Size,stdin);
        tail=(head=buffer)+l;
    }
    if(head==tail) return -;
    return *head++;
}
inline int read() {
    int x=,f=;char c=Getchar();
    for(;!isdigit(c);c=Getchar()) if(c=='-') f=-;
    for(;isdigit(c);c=Getchar()) x=x*+c-'';
    return x*f;
}
const int mod = , maxn = ;
int a[maxn], r[maxn], lg[maxn], n, k;
inline int skr(int x, int t) {
    int res = ;
    while (t) {
        if (t & )
            res = 1LL * res * x % mod;
        x = 1LL * x * x % mod;
        t = t >> ;
    }
    return res;
}
int wn[maxn], iwn[maxn];
void init(int n) {
    wn[] = iwn[] = ;
    rep(i, , n-) wn[i] = skr(, (mod - ) / (i << ));
    rep(i, , n-) iwn[i] = skr(, (mod - ) / (i << ));
}
inline void fft_init(int n) {rep(i, , n - ) r[i] = (r[i >> ] >> ) | ((i & ) << (lg[n] - ));}
inline void fft(int *a, int n, int type) {
    rep(i, , n - ) if (i < r[i]) swap(a[i], a[r[i]]);
    for (int i = ; i < n; i <<= ) {
        //int wn = skr(3, (mod - 1) / (i << 1));
        //if (type == -1)
        //    wn = skr(wn, mod - 2);
        int twn = (type == -) ? iwn[i] : wn[i];
        for (int j = ; j < n; j += (i << )) {
            int w = ;
            for (int k = ; 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 == -) {
        int inv_n = skr(n, mod - );
        rep(i, , n - ) 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, );
    // for(int i=0;i<len;i++)cout<<A[i]<<" ";
    // cout<<endl;
    fft(B, len, );
    for (int i = ; i < len; i++) A[i] = (LL)A[i] * B[i] % mod;
    fft(A, len, -);
    --len;
    while (!A[len]) --len;
    return len;
}
vector<int> poly[maxn];
int solve(int l, int r) {
    if (l == r)
        return poly[l].size() - ;
    int mid = (l + r) >> ;
    int ls = solve(l, mid), rs = solve(mid + , r);
    int L = ;
    for (; L <= ls + rs; L <<= )
        ;

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

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

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