算法步驟

1、用 cur 維護當前和 (初始化為0)，用 res 保存最終結果 (初始化為 -inf)，用 temp 維護開始的位置

2、遍歷 a 數組，cur+=a[i] ，如果 cur < 0，則 cur=0 ,temp=i+1從新開始

3、如果 cur>res，更新res=cur，l=temp，r=i

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 100010;
const int inf = 0x7fffffff;

int n,l,r;
 
int a[maxn];
long long maxSum(){
    long long cur = 0,res=-inf;
    int temp = 1;
    for (int i = 1; i <= n; i++){
        cur += 1ll*a[i];
        if (cur > res){
            res = cur;
            l = temp;
            r = i;
        }
        if (cur < 0){
            cur = 0;
            temp 
 
= i + 1;
        }
    }
    return res;
}
int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];

    cout << maxSum()<<" ";
    cout<< l <<" "<< r << endl;

    return 0;
}

最大字段和