Problem:
Score : 300 points

Problem Statement

In a flower bed, there are N flowers, numbered 1,2,......,N. Initially, the heights of all flowers are 0.
You are given a sequence h=\{h_1,h_2,h_3,......\} as input. You would like to change the height of Flower k to h_k for all k (1 \leq k \leq N), by repeating the following "watering" operation:

Specify integers l and r. Increase the height of Flower x by 1 for all x such that l \leq x \leq r.

Find the minimum number of watering operations required to satisfy the condition.

Constraints

1 \leq N \leq 100

0 \leq h_i \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
h_1 h_2 h_3 ...... h_N

Output

Print the minimum number of watering operations required to satisfy the condition.

Sample Input 1

4
1 2 2 1

Sample Output 1

2

The minimum number of watering operations required is 2.
One way to achieve it is:

Perform the operation with (l,r)=(1,3).

Perform the operation with (l,r)=(2,4).

Sample Input 2

5
3 1 2 3 1

Sample Output 2

5

Sample Input 3

8
4 23 75 0 23 96 50 100

Sample Output 3

221

Solution in Go (1.6) (judge verdict: Accepted)

package main

import (
	"fmt"
)

func main() {
	var N, m int
	fmt.Scan(&N)
	goalList := make([]int, N)
	for i := 0; i < N; i++ {
		fmt.Scan(&goalList[i])
	}
	flowerList := make([]int, N)
	//終端に0をつけたリストを作成
	tmpList := make([]int, N+1)

	for {
		for i := 0; i < N; i++ {
			if flowerList[i] < goalList[i] {
				flowerList[i] = flowerList[i] + 1
				tmpList[i] = 1
			} else {
				tmpList[i] = 0
			}
		}
		for k := 0; k < N+1; k++ {
			if tmpList[k] == 1 && tmpList[k+1] == 0 {
				m++
			}
		}

		if !Contains(tmpList, 1) {
			fmt.Println(m)
			break
		}
	}

}

func Contains(a []int, x int) bool {
	for _, n := range a {
		if x == n {
			return true
		}
	}
	return false
}