25| 丢手绢

代码实现

typedef long long ll;
const int N = 2e5 + 10; // 因为要拼接数组，所以长度翻倍
ll a[N], tmp, ret;
int main()
{
    int n; cin >> n;
    ll circ = 0;
    for (int i = 1; i <= n; i++) 
    {
        cin >> a[i];
        circ += a[i];
        a[i + n] = a[i]; // 拼接数组，处理环形
    }
    int l = 1;
    tmp = 0, ret = 0;
    for (int r = 1; r <= 2 * n; r++) // 遍历到2n
    {
        tmp += a[r];
        // 当当前和超过半周长时，从左边移出元素
        while (tmp > circ / 2)
        {
            tmp -= a[l];
            l++;
        }
        // 更新最大的最近距离
        ret = max(ret, tmp);
        // 防止窗口长度超过n（即绕回一圈以上）
        if (r - l + 1 > n)
        {
            tmp -= a[l];
            l++;
        }
    }
    cout << ret << endl;
}

typedef long long ll;
const int N = 1e5 + 10;
ll a[N], tmp, ret;
int main()
{
	int n; cin >> n;
	ll circ = 0;
	for (int i = 1; i <= n; i++) 
	{
		cin >> a[i];
		circ += a[i];
	}
	int l = 1, r = 1;
	while (r <= n)
	{
		// 入窗口
		tmp += a[r];
		// 判断
		while (tmp * 2 > circ)
		{
			ret = max(ret, circ - tmp);
			// 出窗口
			tmp -= a[l];
			l++; 
		 } 
		 
		// 更新结果
		ret = max(ret, tmp); 
		r++;
	}
	cout << ret << endl;
 }