AcWing 4262. 空调
原题链接
简单
作者:
隐灵
,
2024-04-10 10:11:07
,
所有人可见
,
阅读 3
#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int n;
int b[N];
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i ++ ) scanf("%d", &b[i]);
for (int i = 1; i <= n; i ++ ) //将数组转换成目标温度-当前温度
{
int x;
scanf("%d", &x);
b[i] -= x;
}
for (int i = n + 1; i; i -- ) //因为用的都是b[i],倒着写,防止b[i]被更新
b[i] -= b[i - 1];
int res = 0;
for (int i = 1; i <= n + 1; i ++ )
if (b[i] > 0) //指令数量即为正数的个数
res += b[i];
printf("%d", res);
return 0;
}