51nod 1690 区间求和2

 

长度为 $2$ 的可以直接求，之后就不把 $2$ 当质数了。

考虑一对 $(i,j)$ 会出现在多少个区间里，其中 $|j-i+1|\in \mathbb{P}$

当 $i+j\leq n +1$ 时，一对 $(i,j)$ 会出现在 $[|i-j|+1,i+j]$ 这个区间里面所有质数长度里面，记 $s_i$ 表示 $[1,i]$ 里有多少个质数，那么贡献就是 $a_i \times a_j \times (s_{i+j}-s_{|i-j|})$

当 $i+j > n + 1$ 时，一对 $(i,j)$ 会出现在 $[|i-j|+1,2n-(i+j)-1]$ 这个区间里面所有质数长度里面，贡献即为 $a_i \times a_j \times (s_{2n-(i+j)-1}-s_{|i-j|})$

$|i-j$ 可以把另一个序列翻转一下就变成正常的卷积了。

#include <bits/stdc++.h>
using db = long double;

const int N = (1 << 19) + 1;
const int MOD = 1e9 + 7;
int prime[N], prin, s[N];
bool vis[N];

void M(int &a) {
    if (a >= MOD) a -= MOD;
    if (a < 0) a += MOD;
}

namespace FFT {
    const db pi = acos(-1.0);
    struct Complex {
        db r, i;
        Complex() {}
        Complex(db r, db i): r(r), i(i) {}
        Complex operator + (const Complex &p) const { return Complex(r + p.r, i + p.i); }
        Complex operator - (const Complex &p) const { return Complex(r - p.r, i - p.i); }
        Complex operator * (const Complex &p) const { return Complex(r * p.r - i * p.i, r * p.i + i * p.r); }
    } A[N], B[N];
    int n, l, r[N];
    void init(int m) {
        n = 1, l = 0;
        while (n <= m) n <<= 1, l++;
        for (int i = 0; i < n; i++) {
            r[i] = r[i >> 1] >> 1 | ((i & 1) << (l - 1));
        }
    }
    void FFT(Complex *a, int pd) {
        for (int i = 0; i < n; i++)
            if (i < r[i])
                std::swap(a[i], a[r[i]]);
        for (int mid = 1; mid < n; mid <<= 1) {
            Complex wn(cos(pi / mid), pd * sin(pi / mid));
            for (int l = mid << 1, j = 0; j < n; j += l) {
                Complex w(1.0, 0.0);
                for (int k = 0; k < mid; k++, w = w * wn) {
                    Complex u = a[k + j], v = w * a[k + j + mid];
                    a[k + j] = u + v;
                    a[k + j + mid] = u - v;
                }
            }
        }
        if (pd == -1)
            for (int i = 0; i < n; i++)
                a[i] = Complex(a[i].r / n, a[i].i / n);
    }
    void Mul(Complex *a, Complex *b) {
        FFT(a, 1); FFT(b, 1);
        for (int i = 0; i < n; i++)
            a[i] = a[i] * b[i];
        FFT(a, -1);
    }
    void solve(int *a, int m) {
        FFT::init(m * 2);
        int ans = 0;
        for (int i = 1; i < m; i++)
            M(ans += 2 * a[i] * a[i + 1]);
        for (int i = 1; i <= m; i++)
            A[i] = B[i] = Complex(1.0l * a[i], 0.0l);
        Mul(A, B);
        for (int i = 1; i <= 2 * m; i++) {
            if (i % 2 == 0) {
                int t = (long long)(A[i].r + 0.5l) % MOD;
                if (i <= m + 1)
                    M(ans += 1LL * s[i - 1] * t % MOD);
                else
                    M(ans += 1LL * s[2 * m - i + 1] * t % MOD);
            }
        }
        memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
        for (int i = 1; i <= m; i++) {
            A[i] = Complex(1.0l * a[i], 0.0l);
            B[i] = Complex(1.0l * a[m - i + 1], 0.0l);
        }
        Mul(A, B);
        for (int i = 1; i <= 2 * m; i++) {
            int len = std::abs(i - m - 1);
            if (len % 2 == 0) {
                int t = (long long)(A[i].r + 0.5l) % MOD;
                M(ans -= 1LL * t * s[len] % MOD);
            }
        }
        printf("%d\n", ans);
    }
}

void init(const int n) {
    for (int i = 2; i <= n; i++) {
        if (!vis[i]) prime[++prin] = i, s[i] = 1;
        for (int j = 1; j <= prin && i * prime[j] <= n; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) break;
        }
    }
    s[2] = 0;
    for (int i = 3; i <= n; i++)
        s[i] += s[i - 1];
}

int a[N], n;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", a + i);
    init(2 * n + 100);
    FFT::solve(a, n);
    return 0;
}