題意: 有一個數組a,找得到其(最大區間和sum - 區間內最大元素maxx)的最大值。
思路: 每次遇到這周大範圍的區間和問題就沒轍,其實都是些很巧妙的思維
- 因爲答案ans的最小值爲0,所以只需考慮正整數元素;又因元素max可在[1 ,30]範圍,所以可以枚舉maxx值,再枚舉區間即可。
- 若a[i] > maxx 則不符枚舉目的,sum = 0(放棄前面的區間段)
- 若sum < 0則不符最大值 >= 0,sum = 0
代碼實現:
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;
int n, ans, sum;
int a[N];
signed main()
{
IOS;
cin >> n;
for(int i = 0; i < n; i ++)
cin >> a[i];
//枚舉maxx
for(int maxx = 1; maxx <= 30; maxx ++){
sum = 0;
//再枚舉區間
for(int i = 0; i < n; i ++){
if(a[i] > maxx){
sum = 0;
continue;
}
sum += a[i];
if(sum < 0)
sum = 0;
ans = max(ans, sum - maxx);
}
}
cout << ans << endl;
return 0;
}