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程序世界

#include <iostream>
using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        long long n;
        cin >> n;

        // 数学结论不因语言改变
        if (n % 3 == 0) {
            cout << 2 * n << endl;
        } else {
            cout << n << endl;
        }
    }
    return 0;
}

t = int(input())
while t:
    t -= 1
    n = int(input())

    # 只有当 n 是 3 的倍数，周期排列才能首尾相接
    if n % 3 == 0:
        print(2 * n)
    else:
        print(n)

#include <iostream>
using namespace std;
long long a[200005], sum[200005];

int main() {
    long long n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        sum[i] = sum[i - 1] + a[i]; // 前缀和
    }

    long long s = 0;
    for (int i = 1; i <= n - 1; i++) {
        s += a[i] * (sum[n] - sum[i]); // 右侧之和
    }
    cout << s << endl;
    return 0;
}

n = int(input())
a = [0] + list(map(int, input().split()))
prefix = [0] * (n + 1)

for i in range(1, n + 1):
    # prefix[i] 表示前 i 个数之和
    prefix[i] = prefix[i - 1] + a[i]

answer = 0
for i in range(1, n):
    # 固定 a[i]，一次得到它与右边所有数的乘积和
    answer += a[i] * (prefix[n] - prefix[i])

print(answer)

long long gcd(long long a, long long b) {
    return b > 0 ? gcd(b, a % b) : a;
}

bool isPrime(long long x) {
    for (long long i = 2; i * i <= x; i++) {
        // 找到一个非平凡因数，x 就不是质数
        if (x % i == 0) return false;
    }
    return true;
}

from math import gcd, isqrt

def is_prime(x):
    # 只检查到整数平方根
    for i in range(2, isqrt(x) + 1):
        if x % i == 0:
            return False
    return True

long long w, h;
int main() {
    cin >> w >> h;
    long long x = gcd(w, h);

    if (x == 1) {
        cout << 0 << endl;
        return 0;
    }
    if (isPrime(x)) {
        cout << (w / x) * (h / x) << endl;
        return 0;
    }
    for (long long i = 2; i * i <= x; i++) {
        if (x % i == 0) {
            cout << (w / i) * (h / i) << endl;
            return 0;
        }
    }
}

w, h = map(int, input().split())
x = gcd(w, h)

if x == 1:
    print(0)
elif is_prime(x):
    # x 是质数时，最小合法边长只能是 x
    print((w // x) * (h // x))
else:
    for i in range(2, isqrt(x) + 1):
        if x % i == 0:
            # 第一个因数就是最小合法边长
            print((w // i) * (h // i))
            break