数组的子数组之和的最大值

#include<iostream>
#include <stdio.h>
using namespace std;

//最直接的方法
int MaxSum(int* A, int n) {
    int maximum = 0xffffffff;
    int sum;
    for(int i = 0; i<n;i++) {
        
        for(int j = i; j < n; j++){
            sum = 0;
            for(int k = i; k <=j; k++) {
                sum += A[k];
            } 
            if(sum > maximum)
                maximum = sum;
        }
    }
    return maximum;
}

//二分法递归

int BinarySum(int* A, int beg, int end) {
    if(beg == end)
        return A[beg];
    int mid = (beg + end) / 2;
    int leftSum = BinarySum(A, beg, mid);
    int rightSum = BinarySum(A, mid+1, end);

    int midSum = A[mid+1];
    int sum = A[mid+1];
    for(int i = mid+2; i <= end; i++ ) {
        sum += A[i];
        if(midSum < sum)
            midSum = sum;
    }
    midSum += A[mid];
    sum = midSum;
    for(i = mid-1; i>= beg; i--){
        sum += A[i];
        if(midSum < sum)
            midSum = sum;
    }
    return ( leftSum>rightSum?leftSum:rightSum ) > midSum ? ( leftSum>rightSum?leftSum:rightSum ):midSum;

}

int main() {
    int nums[2] = {1, 5};
    //动态规划
    int maxSum = nums[0];
    int partSum = nums[0];
    for(int i = 1; i < 2; i++) {
        if(partSum < 0)
            partSum = 0;

        partSum += nums[i];
        
        if(partSum > maxSum)
            maxSum = partSum;
    }
    cout << maxSum << endl;

    cout << MaxSum(nums, 2)<<endl;

    cout <<  BinarySum(nums, 0, 1) << endl;
    return 0;
}