codeforces 463B Caisa and Pylons

最新推荐文章于 2017-09-07 16:02:36 发布

风吹过的烟花

最新推荐文章于 2017-09-07 16:02:36 发布

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本文介绍了一道编程挑战题目，玩家需要帮助角色Caisa通过调整电塔高度来完成游戏任务。文章提供了算法思路及实现代码，讨论了如何最小化花费以达到目标。

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Caisa solved the problem with the sugar and now he is on the way back to home.

Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.

Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?

Input

The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.

Output

Print a single number representing the minimum number of dollars paid by Caisa.

   Examples
 
     input
   
5
3 4 3 2 4

     output
   
4

     input
   
3
4 4 4

     output
   
4

Note

In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.

今天小比赛的第一题，竟然错了两次，不该啊！！！

#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
    int ma=0,s=0;
    int n=0,a[100010];
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<n;i++)
    {
        s+=a[i+1]-a[i];
        ma=max(ma,s);
    }
    if(ma>0)
        printf("%d\n",ma+a[1]);
    else
        printf("%d\n",a[1]);
    return 0;
}