Three ways to find the largest sub-segment sum

Note here that the sub-segment is continuous, and the sub-sequence is not continuous.

(1) Violent solution

//最大子段和 暴力 直接三层循环 暴力所有区间 然后找最大的子段和 
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>

using namespace std;
int a[2000];

int main(){
    ios::sync_with_stdio(false);
    cin.tie();
    int n;
    cin >> n;
    for(int i = 0;i < n; i++)
        cin>>a[i];
    int ans =0;
    for(int i = 0;i < n;i++){
        for(int j =i ; j<n ;j++){
            int sum = 0;
            for(int k = i;k<=j;k++){
                sum +=a[k];
            }
            ans=max(sum,ans);
        }
    }


    cout<< ans <<endl;
    return 0;
}

//最大子段和 暴力优化
//a（i~j）可分解为 a(j)+a(i~j-1) 然后发现可以将刚刚的最里层循环给省掉 从而节省一部分时间
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>

using namespace std;
int a[2000];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie();
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
        cin>>a[i];
    int ans =0;
    for(int i = 0; i < n; i++)
    {
        int sum = 0;
        for(int j =i ; j<n ; j++)
        {
            sum +=a[j];
            ans=max(sum,ans);
        }
    }
    cout<< ans <<endl;
    return 0;
}

(2) The idea of ​​using divide and conquer algorithm

a[0:n-1] can be divided into a[0:n/2] and a[n/2+1:n-1], the two ends of a[1:n] are the largest sub-segments and there are the following three types Happening

1. The maximum sub-segment sum of a[0:n-1] is the same as the maximum sub-segment sum of a[0:n/2], which can be solved by recursion

2. The maximum sub-segment sum of a[0:n-1] is the same as the maximum sub-segment sum of a[n/2+1:n-1], which can be solved by recursion

3. The sum of the largest sub-segments of a[0:n-1] is sum(ak) k from i to j 1<=i<=n/2,n/2+1<=j<=n Obviously a[ n/2] and a[n/2+1] are in the optimal subsequence. So you can calculate sl=max (sum(ak)) k from i to n/2 in a[0:n/2]. 0<=i<=n/2 Similarly a[n/2+1:n ] Calculate sr=max (sum(ak)) k from n/2+1 to nn/2+1<=i<=n-1

//最大子段和 分治写法
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>

using namespace std;
int a[2000];
int f(int l,int r){
    int sum = 0;
    if(l==r)
        sum=max(a[l],0);
    else{
        int mid = (floor)((l+r)/2);
        int lsum = f(l,mid);
        int rsum = f(mid+1,r);
        int sl= 0;
        int tmp = 0;
        for(int i = mid; i >= l; i--){
            tmp += a[i];
            sl=max(sl,tmp);
        }
        tmp = 0;
        int sr = 0;
        for(int i=mid + 1;i <= r; i++){
            tmp += a[i];
            sr=max(sr,tmp);
        }
        sum=max(sl+sr,max(lsum,rsum));
    }
    return sum;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie();
    int n;
    cin >> n;
    for(int i = 0;i < n; i++)
        cin>>a[i];
    int ans = f(0,n-1);
    cout<< ans <<endl;
    return 0;
}

(3) dp solution

Take the set of examples and run the dynamic transfer equation directly, and then find the maximum each time

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
#define ll long long

const int N=1e5+6;
ll dp[N];

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin>>n;
    memset(dp,0,sizeof(dp));
    ll maxans = -1;
    int a;
    for(int i=0;i<n;i++){
        cin>>a;
        dp[i]=max(dp[i],dp[i-1]+a);
        maxans=max(maxans,dp[i]);
    }
    cout << maxans << endl;
    return 0;
}