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++14 (GCC 5.4.1) (judge verdict: Accepted)

//#pragma GCC optimize("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#pragma warning(disable:4786)
#pragma warning(disable:4996)
#include <ctime>
#include<list>
#include <numeric>
#include<bitset>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<functional>
#include<string>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<utility>
#include<fstream>
#include<sstream>
#include<cmath>
#include<stack>
#include<assert.h>
#include<unordered_map>
#include<unordered_set>
#include <array>
using namespace std;

#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
	~debug() { cerr << endl; }
	template<class c> typename enable_if<sizeof dud<c>(0) != 1, debug&>::type operator<<(c i) {
		cerr << boolalpha << i; return *this;
	}

	template<class c, int=0> typename enable_if<sizeof dud<c>(0) == 1, debug&>::type operator<<(c i) {
		return *this << range(begin(i), end(i));
	}

sim, class b dor(pair < b, c > d) {
	ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
	*this << "[";
	for (auto it = d.b; it != d.e; ++it)
		*this << ", " + 2 * (it == d.b) << *it;
	ris << "]";
}
#else
	sim dor(const c&) { ris; }
#endif
};
#define watch(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

#define MEM(a, b) memset(a, (b), sizeof(a))
#define CLR(a) memset(a, 0, sizeof(a))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define ABS(X) ( (X) > 0 ? (X) : ( -(X) ) )
#define S(X) ( (X) * (X) )
#define SZ(V) (int )V.size()
#define FORN(i, n) for(int i = 0; i < n; i++)
#define FORAB(i, a, b) for(int i = a; i <= b; i++)
#define ALL(V) V.begin(), V.end()
#define IN(A, B, C)  ((B) <= (A) && (A) <= (C))
#define AIN(A, B, C) assert(IN(A, B, C))

//typedef int LL;
typedef long long int LL;
//typedef __int128 LLL;
typedef long long LLL;

typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef pair<double, double> PDD;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<PLL > VPL;
typedef vector<PII > VP;
typedef vector<double> VD;
typedef vector<vector<int>> VVI;
typedef long double ld;

const LL MOD = 998244353;
int h[102];

int solve(int a, int b) {
	if (a == b) return h[a];
	if (h[a] == 0) return solve(a + 1, b);
	if (h[b] == 0) return solve(a, b - 1);
	int x = 1000;
	for (int i = a; i <= b; i++) {
		if (h[i] == 0) {
			return solve(a, i - 1) + solve(i + 1, b);
		}
		x = MIN(x, h[i]);
	}
	for (int i = a; i <= b; i++) {
		h[i] -= x;
	}
	return x + solve(a, b);
}

void solve(int ks) {
	int n;
	scanf("%d", &n);
	FORN(i, n) scanf("%d", &h[i]);
	int ans = solve(0, n - 1);
	printf("%d\n", ans);
}

int main()
{
#ifdef LOCAL
	double start_time = clock();
	freopen("C:\\Home\\ContestCodes\\sample.in", "r", stdin);
	//	freopen("out.out", "w", stdout);
#endif


	if (0) {
		int T;
		scanf("%d", &T);
		//AIN(T, 1, 25);
		for (int ks = 1; ks <= T; ks++) {
			solve(ks);
			fprintf(stderr, "%d done\n", ks);
		}
	}
	else {
		solve(0);
	}

#ifdef LOCAL
	double end_time = clock();
	fprintf(stderr, "Time = %lf\n", (end_time - start_time) / CLOCKS_PER_SEC);
#endif

	return 0;
}