1<=n<=100000
by manong_czk @ 2024-03-27 14:52:32
显然只跟相邻两项的差有关，剩下自己想去
by jason_sun @ 2024-03-27 14:55:35
@[jason_sun](/user/399762) 感谢已关
by manong_czk @ 2024-03-27 14:57:00
蒟蒻不懂**球代码** $\color{purple}\texttt{Please!}$
by manong_czk @ 2024-03-27 15:07:39
@[manong_czk](/user/1063924)
```cpp
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f3f3f3f3fll;
ll n, a[100005] = {}, d[100005], dmin = inf; // d[i] = a[i] - a[i - 1]，dmin表示d[i]的最小值，即a中相邻两项的差的最小值
int main() {
cin >> n;
for (ll i = 1; i <= n; i++) {
cin >> a[i];
}
for (ll i = 2; i <= n; i++) {
d[i] = a[i] - a[i - 1];
}
for (ll i = 2; i <= n; i++) {
if (d[i] < 0) { // d[i] < 0 => a[i] - a[i - 1] < 0 => a[i] < a[i - 1]
cout << 0 << endl;
exit(0); // 相当于在主函数内执行return 0;
}
dmin = min(dmin, d[i]);
}
cout << (dmin + 2) / 2 << endl;
// 将两数的差值从dmin减到负数，则至少要减(dmin + 1)
// (dmin + 1) / 2 上取整 = ((dmin + 1) + (2 - 1)) / 2 = (dmin + 2) / 2
return 0;
}
```
by 0x28202e202e29 @ 2024-03-27 16:19:39
@[0x28202e202e29](/user/790188) 万般感谢大佬！已关。【叩拜】
by manong_czk @ 2024-03-27 18:13:17
@[0x28202e202e29](/user/790188) 蛋柿
```
5
1 2 3 4 5
```
这个答案应该不是 `1` 吧。
by manong_czk @ 2024-03-27 19:40:14
o，答案是 1。
by manong_czk @ 2024-03-27 19:49:18