C++ 解法, 执行用时: 7ms, 内存消耗: 1624KB, 提交时间: 2020-08-18

#ifdef debug
#include <time.h>
#endif

#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <math.h>

#define MAXN ((int)1e5+7)
#define ll long long int
#define INF (0x7f7f7f7f)
#define fori(lef, rig) for(int i=lef; i<=rig; i++)
#define forj(lef, rig) for(int j=lef; j<=rig; j++)
#define fork(lef, rig) for(int k=lef; k<=rig; k++)
#define QAQ (0)

using namespace std;

#define show(x...) \
    do { \
       cout << "\033[31;1m " << #x << " -> "; \
       err(x); \
    } while (0)

void err() { cout << "\033[39;0m" << endl; }
template<typename T, typename... A>
void err(T a, A... x) { cout << a << ' '; err(x...); }

namespace FastIO{

    char print_f[105];
    void read() {}
    void print() { putchar('\n'); }

    template <typename T, typename... T2>
       inline void read(T &x, T2 &... oth) {
           x = 0;
           char ch = getchar();
           ll f = 1;
           while (!isdigit(ch)) {
               if (ch == '-') f *= -1; 
               ch = getchar();
           }
           while (isdigit(ch)) {
               x = x * 10 + ch - 48;
               ch = getchar();
           }
           x *= f;
           read(oth...);
       }
    template <typename T, typename... T2>
       inline void print(T x, T2... oth) {
           ll p3=-1;
           if(x<0) putchar('-'), x=-x;
           do{
                print_f[++p3] = x%10 + 48;
           } while(x/=10);
           while(p3>=0) putchar(print_f[p3--]);
           putchar(' ');
           print(oth...);
       }
} // namespace FastIO
using FastIO::print;
using FastIO::read;

int n, m, Q, K, a[MAXN], dp[MAXN];

signed main() {
#ifdef debug
    freopen("test.txt", "r", stdin);
    clock_t stime = clock();
#endif
    read(n);
    for(int i=1; i<=n; i++) read(a[i]);
    dp[1] = a[1];
    // for(int i=1; i<=n; i++) show(i, a[i]);
    for(int i=2; i<=n; i++) 
        if(a[i] <= a[i-1]) {
            dp[i] = dp[i-1];
        } else {
            dp[i] = dp[i-1] + (a[i] - a[i-1]);
        }
    // for(int i=1; i<=n; i++) show(dp[i]);
    printf("%d\n", dp[n]);




#ifdef debug
   clock_t etime = clock();
   printf("rum time: %lf 秒\n",(double) (etime-stime)/CLOCKS_PER_SEC);
#endif 
   return 0;
}

C++ 解法, 执行用时: 8ms, 内存消耗: 1272KB, 提交时间: 2020-07-28

#include<bits/stdc++.h>
using namespace std;
  
typedef long long LL;
int n,a[100005];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int ans=0;
    cin>>n;
    for(int i=1;i<=n;++i)
        cin>>a[i];
    for(int i=1;i<=n;++i)
    {
        if(a[i]<=a[i+1])//如果前一个位置比后一个位置小，则种后一个的同时把前一个种了
            continue;
        ans+=a[i]-a[i+1];//如果前一个位置比后一个位置大，先把前一个种a[i]-a[i+1]
    }
    cout<<ans<<endl;
    return 0;
}