题目传送门

解题思路：

点这里

AC代码：

 1 #include<iostream>
 2 #include<cstdio>
 3 
 4 using namespace std;
 5 
 6 int n,a[100001],u,o;
 7 
 8 int main() {
 9     scanf("%d",&n);
10     for(int i = 1;i <= n; i++)
11         scanf("%d",&a[i]);
12     o = a[1];
13     for(int i = 2;i <= n; i++) {
14         u = o;
15         o= a[i];
16         a[i] = o - u;
17     }
18     u = 0;
19     for(int i = 1;i <= n ;i++) 
20         if(a[i] > 0) u += a[i];
21     printf("%d",u);
22     return 0;
23 }

//NOIP提高 2018 Day1 T1

标签：洛谷,int,Day1,传送门,铺设,printf,P5019,include
来源： https://www.cnblogs.com/lipeiyi520/p/11267260.html