Description
定义二元运算 opt 满足
现在给定一个长为 n 的数列 a 和一个长为 m 的数列 b ，接下来有 q 次询问。每次询问给定一个数字 c
你需要求出有多少对 (i, j) 使得 a_i opt b_j=c 。
Input
第一行是一个整数 T (1≤T≤10) ，表示测试数据的组数。
对于每组测试数据：
第一行是三个整数 n,m,q (1≤n,m,q≤50000) 。
第二行是 n 个整数，表示 a_1,a_2,?,a_n (0≤a_1,a_2,?,a_n≤50000) 。
第三行是 m 个整数，表示 b_1,b_2,?,b_m (0≤b_1,b_2,?,b_m≤50000) 。
第四行是 q 个整数，第 i 个整数 c_i (0≤c_i≤100000) 表示第 i 次查询的数。
Output
对于每次查询，输出一行，包含一个整数，表示满足条件的 (i, j) 对的个数。
Sample Input
2
2 1 5 1 3 2 1 2 3 4 5 2 2 5 1 3 2 4 1 2 3 4 5Sample Output
1
0 1 0 0 1 0 1 0 1思路
在上面两种情况中，减法可以把y取反直接fft，然后加法因为有限制可以在值域上进行分治fft
然后其实很板子的。。。
#includeusing namespace std; typedef long long ll; namespace io { const int BUFSIZE = 1 << 20;char ibuf[BUFSIZE], *is = ibuf, *it = ibuf;char obuf[BUFSIZE], *os = obuf, *ot = obuf + BUFSIZE - 1; char read_char() { if (is == it) it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin); return *is++;} int read_int() { int x = 0, f = 1; char c = read_char(); while (!isdigit(c)) { if (c == '-') f = -1; c = read_char(); } while (isdigit(c)) x = x * 10 + c - '0', c = read_char(); return x * f;} ll read_ll() { ll x = 0, f = 1; char c = read_char(); while (!isdigit(c)) { if (c == '-') f = -1; c = read_char(); } while (isdigit(c)) x = x * 10 + c - '0', c = read_char(); return x * f;} void read_string(char* s) { char c = read_char(); while (isspace(c)) c = read_char(); while (!isspace(c)) *s++ = c, c = read_char(); *s = 0;} void flush() { fwrite(obuf, 1, os - obuf, stdout); os = obuf;} void print_char(char c) { *os++ = c; if (os == ot) flush();} void print_int(int x) { static char q[20]; if (!x) print_char('0'); else { if (x < 0) print_char('-'), x = -x; int top = 0; while (x) q[top++] = x % 10 + '0', x /= 10; while (top--) print_char(q[top]); }} void print_ll(ll x) { static char q[20]; if (!x) print_char('0'); else { if (x < 0) print_char('-'), x = -x; int top = 0; while (x) q[top++] = x % 10 + '0', x /= 10; while (top--) print_char(q[top]); }} struct flusher_t { ~flusher_t() { flush(); }} flusher; };using namespace io; const int N = 3e5 + 10;const int M = 5e4 + 10;const double eps = 1e-6;const double PI = acos(-1); typedef complex Complex;typedef vector Poly; Complex w[2][N]; void init() { for (int i = 1; i < (1 << 18); i <<= 1) { w[0][i] = w[1][i] = Complex(1, 0); Complex wn(cos(PI / i), sin(PI / i)); for (int j = 1; j < i; j++) w[1][i + j] = w[1][i + j - 1] * wn; wn = Complex(cos(PI / i), -sin(PI / i)); for (int j = 1; j < i; j++) w[0][i + j] = w[0][i + j - 1] * wn; }} void transform(Complex *t, int len, int typ) { for (int i = 0, j = 0, k; j < len; j++) { if (i > j) swap(t[i], t[j]); for (k = (len >> 1); k & i; k >>= 1) i ^= k; i ^= k; } for (int i = 1; i < len; i <<= 1) { for (int j = 0; j < len; j += (i << 1)) { for (int k = 0; k < i; k++) { Complex x = t[j + k], y = w[typ][i + k] * t[j + k + i]; t[j + k] = x + y; t[j + k + i] = x - y; } } } if (typ) return; for (int i = 0; i < len; i++) t[i] = Complex(t[i].real() / (double) len, t[i].imag());} bool equ0(double x) { return fabs(x) < eps;} Poly mul(const Poly a, const Poly b) { static Poly cura, curb; cura = a, curb = b; int len = 1 << (int) ceil(log2(cura.size() + curb.size() - 1)); cura.resize(len), curb.resize(len); transform(&cura[0], len, 1); transform(&curb[0], len, 1); for (int i = 0; i < len; i++) cura[i] *= curb[i]; transform(&cura[0], len, 0); return cura;} Poly ansadd(N, 0), anssub(N, 0);int n, m, q, maxa, maxb;int a[N], b[N];int cnta[N], cntb[N]; void devide(int l, int r) { if (l == r) return; int mid = (l + r) >> 1; devide(l, mid); devide(mid + 1, r); Poly x, y; int sizl = mid - l + 1, sizr = r - mid; x.resize(sizl); y.resize(sizr); for (int i = 0; i < sizl; i++) x[i] = cnta[i + l]; for (int i = 0; i < sizr; i++) y[i] = cntb[i + mid + 1]; x = mul(x, y); for (int i = 0; i < (signed) x.size(); i++) ansadd[i + l + mid + 1] += x[i];} void calcadd() { devide(0, max(maxa, maxb));} void calcsub() { static Poly x, y; x.resize(maxb + maxa + 1); y.resize(maxb + maxa + 1); for (int i = 0; i <= maxb + maxa; i++) x[i] = y[i] = 0; for (int i = 0; i <= maxa; i++) x[maxb + i] = cnta[i]; for (int i = 0; i <= maxb; i++) y[maxb - i] = cntb[i]; x = mul(x, y); for (int i = maxb * 2; i < (signed) x.size(); i++) anssub[i - maxb * 2] = x[i];} void solve() { n = read_int(), m = read_int(), q = read_int(); maxa = maxb = 0; for (int i = 1; i <= n; i++) { cnta[a[i] = read_int()]++; maxa = max(maxa, a[i]); } for (int i = 1; i <= m; i++) { cntb[b[i] = read_int()]++; maxb = max(maxb, b[i]); } calcsub(); calcadd(); while (q--) { int c = read_int(); print_ll((ll) round(ansadd[c].real()) + (ll) round(anssub[c].real())); print_char('\n'); } for (int i = 0; i <= maxa; i++) cnta[i] = 0; for (int i = 0; i <= maxb; i++) cntb[i] = 0; for (int i = 0; i <= maxa + maxb; i++) ansadd[i] = anssub[i] = 0;} int main() {#ifdef dream_maker freopen("input.txt", "r", stdin);#endif init(); int T = read_int(); while (T--) solve(); return 0;}