結果
| 問題 | 
                            No.1120 Strange Teacher
                             | 
                    
| コンテスト | |
| 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 | 
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| 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_pair
using 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 > 0
    ll 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;
}