【CODEFORCES】 B. Caisa and Pylons

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 h_i. 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 h_k - h_k + 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, ..., h_n (1  ≤  h_i  ≤  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.

题解：题意很好懂，就是找出H[I]和H[I-1]的最小差值就行了。然后判断如果大于等于零就说明不用额外的能量，答案就是H[1]的高度，否则答案就是 H[1]-最小差值 。    

水水更健康~~~~~

#include <iostream>
#include <cstdio>

using namespace std;

int n,i;  long long h[100002],ans,t;
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        scanf("%I64d",&h[i]);
    ans=0; t=0;
    for (int i=1;i<=n-1;i++)
    {
        t+=h[i]-h[i+1];
        ans=min(ans,t);
    }
    if (ans>=0) ans=h[1];
    else ans=h[1]-ans;
    cout <<ans<<endl;
    return 0;
}