题目链接点击这里。
思路
此题分奇偶两种情况进行讨论:
- 当n为偶数时,很容易证明最优解就是两个相邻的数的和最大;
- 当n为奇数时,不太好处理
我们想,当n为奇数时,若假设这时候p个礼物满足条件,第一个人将p个礼物分为1,…,r1和r1,…,p这两种情况,第i个人若为奇数,则要尽量往r1,…,p取,同样,若第i个人为偶数,则要尽量往1,…r1取,这样可以使得最后一个人(奇数)能取到尽可能多的数。
实现
- 由于此题只需要输出最优解,因此我们维护两个数组
Right
和Left
,分别表示第i个人在两个范围内的取值情况,依次递归即可检验p个礼物是否可行。 - 很容易验证p的最大值为礼物最大数的3倍,因此可以使用二分法减少迭代次数。
#include <iostream>
using namespace std;
int n;
const int maxn = 1000000 + 10;
int guard[maxn], Left[maxn], Right[maxn];
int test(int p)
/* Left = [1,...,r1]
Right = [r1+1,...p]
*/
{
int x = guard[1], y = p - guard[1];
Left[1] = x;
Right[1] = 0;
for (int i = 2 ;i <= n; i++)
{
if (i % 2)
/* if odd */
{
/* get Right side as much as posible, except those token by i-1 */
Right[i] = min(y - Right[i - 1], guard[i]);
Left[i] = guard[i] - Right[i];
}
else
/* if even */
{
/* get Left side as much as posible, except those token by i -1 */
Left[i] = min(x - Left[i - 1], guard[i]);
Right[i] = guard[i] - Left[i];
}
}
/* test Left side of n is zero,if zero, p is ok */
return Left[n] == 0;
}
int main()
{
while (cin >> n && n)
{
for (int i = 1; i <= n; i++)
{
cin >> guard[i];
}
guard[n + 1] = guard[1];
if (n == 1)
{
cout << guard[1] << endl;
continue;
}
int R = 0, L = 0;
for (int i = 1; i <= n; i++)
L = max(L, guard[i + 1]+guard[i]);
if (n % 2)
{
for (int i = 1; i <= n; i++)
R = max(R, guard[i] * 3);
while (L < R)
{
int middle = (R + L) / 2;
if (test(middle))
R = middle;
else
L = middle + 1;
}
}
cout << L << endl;
}
return 0;
}