結果
問題 | No.1120 Strange Teacher |
ユーザー |
|
提出日時 | 2022-04-10 11:55:46 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 70 ms / 1,000 ms |
コード長 | 2,696 bytes |
コンパイル時間 | 3,779 ms |
コンパイル使用メモリ | 229,724 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-11-30 20:28:58 |
合計ジャッジ時間 | 7,421 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 27 |
ソースコード
#include <bits/stdc++.h>#include <atcoder/all>typedef long long int ll;typedef long long int ull;#define MP make_pairusing namespace std;using namespace atcoder;typedef pair<ll, ll> P;const ll MOD = 998244353;// const ll MOD = 1000000007;using mint = modint998244353;// using mint = modint1000000007;const double pi = 3.1415926536;const int MAX = 2000003;long long fac[MAX], finv[MAX], inv[MAX];void COMinit() {fac[0] = fac[1] = 1;finv[0] = finv[1] = 1;inv[1] = 1;for (int i = 2; i < MAX; i++){fac[i] = fac[i - 1] * i % MOD;inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;finv[i] = finv[i - 1] * inv[i] % MOD;}}// 二項係数計算long long COM(int n, int k){if (n < k) return 0;if (n < 0 || k < 0) return 0;return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;}ll gcd(ll x, ll y) {if (y == 0) return x;else if (y > x) {return gcd (y, x);}else return gcd(x % y, y);}ll lcm (ll x, ll y) {return x / gcd(x, y) * y;}ll my_sqrt(ll x) {ll m = 0;ll M = 3000000001;while (M - m > 1) {ll now = (M + m) / 2;if (now * now <= x) {m = now;}else {M = now;}}return m;}ll keta(ll n) {ll ret = 0;while (n) {n /= 10;ret++;}return ret;}ll ceil(ll n, ll m) {// n > 0, m > 0ll ret = n / m;if (n % m) ret++;return ret;}ll pow_ll(ll x, ll n) {if (n == 0) return 1;if (n % 2) {return pow_ll(x, n - 1) * x;}else {ll tmp = pow_ll(x, n / 2);return tmp * tmp;}}int main() {int n;cin >> n;ll a[n];ll b[n];ll asum = 0;ll bsum = 0;for (int i = 0; i < n; i++) {cin >> a[i];asum += a[i];}for (int i = 0; i < n; i++) {cin >> b[i];bsum += b[i];}ll dif = asum - bsum;if (n == 2) {if (dif != 0) {cout << -1 << endl;}else {cout << abs(a[0] - b[0]) << endl;}return 0;}if (dif < 0) {cout << -1 << endl;return 0;}if (dif % (n - 2)) {cout << -1 << endl;return 0;}int op = dif / (n - 2);for (int i = 0; i < n; i++) {a[i] -= op;}int num = 0;for (int i = 0; i < n; i++) {if ((b[i] - a[i] < 0) || ((b[i] - a[i]) % 2)) {cout << -1 << endl;return 0;}num += (b[i] - a[i]) / 2;}if (num == op) {cout << op << endl;}else {cout << -1 << endl;}return 0;}