Description
Statements
Most of the students of the law school prefer visiting photo club to the competitions in Roman law. Members of the photo club visit different interesting places, take photos of each other in front of them, and then rate their photos.
Once they appeared on a unbelievably long street which had n buildings in a row. Every member of the photo club took a photo contained, besides the members of the club and people passing by, a segment of the street. In other words, if you number the buildings in the order they are located on the street, each photo contained some buildings with the consecutive numbers.
Some day a Roman law professor of that law school came across the exhibition of the photos from that street. He hasn't remembered how many photos were there, but he has noticed that the i-th building was captured on ai photos. Now he wants to estimate the minimal number of his students in the photo club, considering that no one could present more that one photo at the exhibition.
Input
The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of buildings.
The second line contains n space-separated integers: ai (0 ≤ ai ≤ 109) — the number of photos that contain the i-th building.
Output
Output a single integer — the minimal number of students in the photo club.
Sample Input
4 1 3 2 0
3
6 1 2 3 1 2 3
5
规律题，直接加上递增数差
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
ll n,a[200005],t,re = 0;
int main()
{
memset(a,0,sizeof(a));
cin >> n;
for(int i=1;i<=n;i++)
{
cin >> a[i];
}
re += a[1];
for(int i=2;i<=n;i++)
{
if(a[i]>a[i-1])
{
re += a[i]-a[i-1];
}
}
cout << re <<endl;
return 0;
}