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1.09 kB
| using namespace std; | |
| using LL = long long; | |
| const int LIM = 500000; | |
| const int MOD = 1000000007; | |
| inline LL add(LL a, LL b) { | |
| return ((a % MOD) + (b % MOD)) % MOD; | |
| } | |
| inline LL sub(LL a, LL b) { | |
| return (((a - b) % MOD) + MOD) % MOD; | |
| } | |
| inline LL mul(LL a, LL b) { | |
| return (a % MOD)*(b % MOD) % MOD; | |
| } | |
| int solve() { | |
| int N, Q; | |
| LL A, B, X, Y; | |
| LL sumA = 0, sumAsq = 0; | |
| LL sumB = 0, sumBsq = 0; | |
| cin >> N; | |
| for (int i = 0; i < N; i++) { | |
| cin >> A >> B; | |
| sumA = (sumA + A) % MOD; | |
| sumB = (sumB + B) % MOD; | |
| sumAsq = (sumAsq + (A*A % MOD)) % MOD; | |
| sumBsq = (sumBsq + (B*B % MOD)) % MOD; | |
| } | |
| cin >> Q; | |
| LL ans = 0; | |
| for (int i = 0; i < Q; i++) { | |
| cin >> X >> Y; | |
| // sum_{j} ((X - A_j)^2) = sum(X^2) - 2*X*sum(A_j) + sum(A_j^2) | |
| ans = add(ans, add(sub(mul(mul(X, X), N), mul(2LL*X, sumA)), sumAsq)); | |
| ans = add(ans, add(sub(mul(mul(Y, Y), N), mul(2LL*Y, sumB)), sumBsq)); | |
| } | |
| return ans; | |
| } | |
| int main() { | |
| int T; | |
| cin >> T; | |
| for (int t = 1; t <= T; t++) { | |
| cout << "Case #" << t << ": " << solve() << endl; | |
| } | |
| return 0; | |
| } | |