AtCoder Beginner Contest 116 C题【题意：可以在任意区间【L,R】上加1，求通过最少加1次数得到题目给定的区间】】{思维好题}

C - Grand Garden

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.

Input

4
1 2 2 1

Output

Input

8
4 23 75 0 23 96 50 100

Output

题意：可以任意在区间【L,R】上加1，求通过最少次数得到题目给定的区间的值】

AC代码：

#include<bits/stdc++.h>

using namespace std;
#define int long long
int arr[152052];
signed  main(){
     int ans=0;
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
        scanf("%lld",&arr[i]);
    int temp=arr[1];
    for(int i=1;i<=n;i++){
        if(i==n){  // 特判一下
            ans+=max(arr[i],temp);
        }else{
            if(arr[i]>=temp){
                temp=arr[i];
            }else{
                ans+=abs(arr[i]-temp);// 需要减去多加的数
                temp=arr[i];
            }
        }
    } 
    cout<<ans;
    return 0;
}

代码2

#include<bits/stdc++.h>

using namespace std;
#define N 515155
int arr[N];

int main(){
  int n;
  cin>>n;
  int ans=0;
  scanf("%d",&arr[1]);
  if(n==1){
    cout<<arr[1];
    return 0;
  }
  int temp=arr[1];
  for(int i=2;i<=n;i++){
    scanf("%d",&arr[i]);
    if(i==n){
      ans+=max(temp,arr[i]);
      break;
    }
    if(arr[i]<temp){
      ans+=temp-arr[i];
      temp=arr[i];
    }else{
      temp=arr[i];
    }
  }
  cout<<ans; 
  return 0;
}