思路:
序列中,只要卖出能赚钱,就记录,求和
题面:
链接:https://ac.nowcoder.com/acm/contest/625/I
来源:牛客网
代码:
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long int
using namespace std;
long long n,a[500005],l;
int main()
{
scanf("%lld",&n);
a[0]=0x3f3f3f3f;
long long ans=0;
for(int i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
if(a[i]>a[i-1])
ans+=a[i]-a[i-1];
}
printf("%lld\n",ans);
return 0;
}