算法导论——求最大子数组问题

#include<iostream>
#include<stdio.h>
#define _CRT_SECURE_NO_WARNINGS
#pragma warning(disable:4996)
int const N = 100;
using namespace std;

struct ARRAY {
    int high;
    int low;
    int sum;
};


int A[N];

ARRAY FIND_MAX_SUBARRAY(int* A, int low, int high);
ARRAY FIND_MAX_CROSSING_SUBARRAY(int* A, int low, int mid, int high);


int main() {
    int i;
    int n=0;
    for (i = 1; i <= N-1; i++) {
        cin >> A[i];
        n++;
        char c;//= cin.get();
        c = getchar();
        if (c != ‘ ‘) break;
    }

    printf("\n");
    ARRAY array;
    array= FIND_MAX_SUBARRAY(A, 1, n);
    for (i = array.low; i <= array.high; i++) {
        printf("%d ", A[i]);
    }
    return 0;
}


ARRAY FIND_MAX_CROSSING_SUBARRAY(int* A, int low, int mid, int high) {
    int left_sum = -999;
    int right_sum = -999;
    int sum = 0;
    int max_left=0,max_right=0;

    int i;
    for (i = mid; i >= low ; i--) {
        sum = sum + A[i];
        if (sum > left_sum) {
            left_sum = sum;
            max_left = i;
        }

    }
    sum = 0;
    for (i = mid+1; i <= high; i++) {
        sum = sum + A[i];
        if (sum > right_sum) {
            right_sum = sum;
            max_right = i;
        }
    }
    ARRAY p;
    p.sum = left_sum + right_sum;
    p.low = max_left;
    p.high = max_right;
    return p;
}


ARRAY FIND_MAX_SUBARRAY(int* A, int low, int high) {
    if (low == high) {
        ARRAY p;
        p.sum = A[low];
        p.low = A[low];
        p.high = A[high];
        return p;
    }
    else {
        ARRAY cross;
        ARRAY left;
        ARRAY right;
        int mid = (low + high) / 2;
        left = FIND_MAX_SUBARRAY(A, low, mid);
        right = FIND_MAX_SUBARRAY(A, mid+1, high);
        cross = FIND_MAX_CROSSING_SUBARRAY(A, low, mid, high);
        if (left.sum >= right.sum && left.sum >= cross.sum) {
            return left;
        }
        else if (left.sum < right.sum && right.sum >= cross.sum) {
            return left;
        }
        else return cross;

    }



    
}