Codeforces Round #264 (Div. 2) B

题目：

B. Caisa and Pylons

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 0has 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.

Sample test(s)

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.

题意分析：

题意：依次走过n+1个地方，每次从第i个走到第i+1个都会有hi-hi+1的能量，正的话，是加血，负的话是扣血，为了确保玩这个游戏，所以要保证血是正的，也可以通过花钱来增加高度来抵消血的损失，求最少的花费。单纯模拟一下就OK，感觉这题一个作为A题的

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define REP(i, s, t) for(int i=s;i<=t;++i)
#define MAXN 100000

typedef long long LL;
int h[MAXN + 5];

int main()
{
    int n;
    LL life = 0;
    LL pay = 0;
    cin >> n;
    REP(i, 1, n)
    cin >> h[i];
    REP(i, 0, n - 1)
    {
        int val = h[i] - h[i + 1];
        if (val < 0)
        {
            if (life + val < 0)
            {
                pay += -(life + val);
                life = 0;
            }
            else
                life += val;
        }
        if (val > 0)
            life += val;
    }
    cout << pay << endl;
    return 0;
}