BZOJ 4836: [Lydsy1704月赛]二元运算

给定的数组不好操作，把他弄成桶，存出现的次数。

然后我们发现如果都是$x+y$或者都是$x-y$就是一个卷积，用FFT就行了。

那考虑分治，对于$[l,mid]$和$[mid+1,r]$两段。

要么左区间选$a$数组，要么右区间选。

然后递归处理即可。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
inline int read() {
    char ch = getchar(); int x = 0, f = 1;
    while(ch < '0' || ch > '9') {
        if(ch == '-') f = -1;
        ch = getchar(); 
    }
    while('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
int aa[50010], bb[50010];
LL cnt[200010];
struct com{
    double r, i;
    com(){
        r = i = 0;
    }
    com(double a, double b) {
        r = a, i = b;
    }
    com operator + (const com& x) const {
        return com(r + x.r, i + x.i);
    }
    com operator - (const com& x) const {
        return com(r - x.r, i - x.i);
    }
    com operator * (const com& x) const {
        return com(r * x.r - i * x.i, r * x.i + i * x.r);
    }
} A[200010], B[200010], w[200010];
const double PI = acos(-1.0);
int rev[200010];
inline void FFT(com a[], int n) {
    for(int i = 0; i < n; ++ i) {
        if(rev[i] > i) {
            swap(a[rev[i]], a[i]);
        }
    }
    for(int t = n >> 1, d = 1; d < n; d <<= 1, t >>= 1) {
        for(int i = 0; i < n; i += (d << 1)) {
            for(int j = 0; j < d; ++ j) {
                com tmp = w[t * j] * a[i + j + d];
                a[i + j + d] = a[i + j] - tmp;
                a[i + j] = a[i + j] + tmp;
            }
        }
    }
}
inline int Do(int n, int m) {
    int L = 0, N = 1;
    for(; N <= n + m; N <<= 1, ++ L);
    for(int i = 0; i < N; ++ i) {
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (L - 1));
        w[i] = com(cos(2.0 * i * PI / N), sin(2.0 * i * PI / N));
    }
    FFT(A, N); FFT(B, N);
    for(int i = 0; i < N; ++ i) {
        A[i] = A[i] * B[i];
        w[i].i = -w[i].i;
    }
    FFT(A, N);
    return N;
}
inline void solve(int l, int r) {
    if(l == r) {
        cnt[0] += 1ll * aa[l] * bb[l];
        return;
    }
    int mid = (l + r) / 2;
    for(int i = l, j = 0; i <= mid; ++ i, ++ j) {
        A[j] = com(aa[i], 0);
    }
    for(int i = mid + 1, j = 0; i <= r; ++ i, ++ j) {
        B[j] = com(bb[i], 0);
    }
    int N = Do(mid - l + 1, r - mid);
    for(int i = 0; i <= r - l - 1; ++ i) {
        cnt[l + mid + 1 + i] += (LL)(A[i].r / N + 0.5);
    }
    for(int i = 0; i < N; ++ i) {
        A[i] = com(0, 0);
        B[i] = com(0, 0);
    }
    for(int i = mid + 1, j = 0; i <= r; ++ i, ++ j) {
        A[j] = com(aa[i], 0);
    }
    for(int i = mid, j = 0; i >= l; -- i, ++ j) {
        B[j] = com(bb[i], 0);
    }
    N = Do(r - mid, mid - l + 1);
    for(int i = 0; i <= r - l - 1; ++ i) {
        cnt[i + 1] += (LL)(A[i].r / N + 0.5);
    }
    for(int i = 0; i < N; ++ i) {
        A[i] = com(0, 0);
        B[i] = com(0, 0);
    }
    solve(l, mid); solve(mid + 1, r);
}
inline void solve() {
    int n = read(), m = read(), q = read();
    memset(aa, 0, sizeof(aa));
    memset(bb, 0, sizeof(bb));
    memset(cnt, 0, sizeof(cnt));
    for(int i = 1; i <= n; ++ i) {
        ++ aa[read()];
    }
    for(int i = 1; i <= m; ++ i) {
        ++ bb[read()];
    }
    solve(0, 50000);
    while(q --) {
        int x = read();
        printf("%lld
", cnt[x]);
    }
}
int main() {
    int T = read();
    while(T --) {
        solve();
    }
}