#include<cstdio>
#include<cstdlib>
#include<cstring>
#define maxn 10020
using namespace std;
int a,d;
int b[maxn];
int main()
{scanf("%d",&a);
for (int i=0;i<a;i++)
scanf("%d",&b[i]);d=b[0];
for (int i=1;i<a;i++)
if (b[i]>=b[i-1]) d=d+b[i]-b[i-1];
printf("%d\n",d);
}

#include<iostream>
#include <cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
int h[100001]={};
int main(){
int n;
cin>>n;

for(int i=1;i<=n;i++){
cin>>h[i];
}
h[0]=0;
int j=0;

for(int i=1;i<=n;i++){
if (h[i]>h[i-1]){j+=h[i]-h[i-1];}

}

cout<<j;

}

VAR
ANS,I,J,M,N:LONGINT;
BEGIN
READLN(N);
FOR I:=1 TO N DO
BEGIN
READ(J);
IF J>M THEN ANS:=ANS+J-M;
M:=J;
END;
WRITELN(ANS);
END.

#include <iostream>
#include <cstdlib>

using namespace std;

int a[100005] = {};
int main(){
int n;
cin >> n;
for(int i = 1; i <= n; ++i){
cin >> a[i];
}
int js = 0;
for(int i = 1; i <= n; ++i){
if(a[i] > a[i - 1]){
js += a[i] - a[i - 1];
}
}
cout << js;
}

###评测结果
编译成功

测试数据 #0: Accepted, time = 15 ms, mem = 476 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 472 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 476 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 472 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 476 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 480 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 476 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 476 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 472 KiB, score = 10
测试数据 #9: Accepted, time = 31 ms, mem = 476 KiB, score = 10
Accepted, time = 61 ms, mem = 480 KiB, score = 100
###代码
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define MAXN 100000+10

int n,first,next,ans;
int main() {
scanf("%d",&n);
scanf("%d",&first);
ans = first;
for(int i = 2; i<=n;i++) {
scanf("%d",&next);
if(first<=next) {
ans+=next-first;
first = next;
}else if(first > next) {
first = next;
}
}
printf("%d",ans);
return 0;
}

#include<stdio.h>
int n,i,j,ans=0,min=21000000,w,min2;
int a[100001]={0};
int main()
{
scanf("%d",&n);
a[0]=0;
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(i=1;i<=n;i++)

if(a[i]>a[i-1]) ans=ans+(a[i]-a[i-1]);
printf("%d",ans);
return 0;
}

#include<cstdio>
int n,A=0,B,ans=0;
int main(){
//freopen("P1844.in","r",stdin);
//freopen("P1844.out","w",stdout);
scanf("%d\n",&n);
for(int i=0;i<n;i++){
scanf("%d ",&B);
if(A<B){
ans+=B-A;
}else;
A=B;
}
printf("%d\n",ans);
}
248KB+77MS

var
a:array[0..100000] of longint;
n,ans,i,t:longint;
begin
readln(n);
for i:=1 to n do
read(a[i]);
ans:=a[1];
t:=a[1];
for i:=2 to n do
begin
if a[i]>t then ans:=ans+a[i]-t;
t:=a[i];
end;
writeln(ans);
end.
水题一道。
真奇怪竟然是在提高组里出的。
这适合普及第二道。
本来还以为我的方法有反例，特地开了个数组，以防不测（可以多扫几遍）。
后来发现一遍过。
今天在转C++，发现真的和Pascal有很大的不同。
好怀念啊。

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 4476 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 4480 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 4476 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 4480 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 4480 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 4480 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 4476 KiB, score = 10
测试数据 #7: Accepted, time = 31 ms, mem = 4480 KiB, score = 10
测试数据 #8: Accepted, time = 125 ms, mem = 4480 KiB, score = 10
测试数据 #9: Accepted, time = 265 ms, mem = 4476 KiB, score = 10
Accepted, time = 421 ms, mem = 4480 KiB, score = 100