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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(){
  int N;
  int h[101]={};//高さのゴール
  scanf("%d",&N);
  int i;
  for(i=1;i<=N;i++) scanf("%d",&h[i]);
  int h_now[101]={};
  int L=1,R=1;
  int count=0;

  while(1){
    //Lを決定する．
    while(L<N){
      if(h_now[L]<h[L]) break;
      else if(h_now[L]==h[L]) L++;
    }
    //printf("L=%d\n",L);

    //Rを決定する.
    R=L;
    while(R<N){
      if(h_now[R+1]<h[R+1]) R++;
      else if(h_now[R+1]==h[R+1]) break;
    }
    //printf("R=%d\n",R);

    //終了条件
    if(L==N && h_now[L]==h[L]) break;

    //水をやる
    for(i=L;i<=R;i++){
      h_now[i]++;
    }
    count++;
  }
  printf("%d\n",count);
  return 0;
}