分治法求解最大子数组问题

/*
最大子数组问题
给出每天股票的价格，求出买进和卖出的时间，使得获利最高。
输入: P[0~n-1]
输出: 买进的时间i和卖出的时间j(0<=i<=j<=n-1)
*/
 
//分治法求解，将数组P转换为数组A，其中A中每个元素A[i]=P[i]-p[i-1]，表示第i-1天买进，第i天卖出的获利。
//那么第i天买进第j天卖出的获利可以表示为A[i+1]+...+A[j],0<=i<j<n
//经过这样的转换后，问题就成了寻找A中最大的非空连续数组，称为最大子数组。
 
//利用分治法求解
//当数组A的长度为1时，A[left,right],left=right,那么i=left-1,j=right;
//当数组A的长度大于1时，将数组A一分为二：m=(left+right)/2;
//最大子数组存在的位置有三个位置：1.数组A[left,m]中；2.数组A[m+1,right]中；3.横跨数组A[left,m]和A[m+1,right]，即i在left~m范围，j在m+1~right范围。
//
 
#include <stdio.h>
#include <stdlib.h>
 
struct ans {
    int low, high, sum;
};
void FindMaxSub(int A[], int low, int high, struct ans *a);
void FindMaxCrosSub(int A[], int low, int mid, int high, struct ans *a);
 
int main()
{
    int *A = NULL;
    int B[17] = { 100, 113, 110, 85, 105, 102, 86, 63, 81, 101, 94, 106, 101, 79, 94, 90, 97 };
    int n = 0, i = 0;
    struct ans Ans;
    scanf("%d", &n);
    A = (int *)malloc(sizeof(int)*n);
    for(i = 0; i<n; i++){
        scanf("%d",A+i);
    }
    for (i = n - 1; i>0; i--) {
        A[i] = A[i] - A[i - 1];
    }
 
    FindMaxSub(A, 1, n - 1, &Ans);
    printf("\n low = %d, high = %d, sum = %d\n", Ans.low, Ans.high, Ans.sum);
 
    system("pause");
    return 0;
 
 
}
 
void FindMaxSub(int A[], int low, int high, struct ans *a)
{
    if (low == high) {
        a->low = a->high = low;
        a->sum = A[low];
    }
    else {
        struct ans a1, a2, a3;
        int mid = (low + high) / 2;
 
        FindMaxSub(A, low, mid, &a1);
        FindMaxSub(A, mid + 1, high, &a2);
        FindMaxCrosSub(A, low, mid, high, &a3);
        // printf("low=%d,mid=%d,high=%d\n", low, mid, high);
        // printf("a1.low=%d, a1.high=%d, a1.sum=%d\n", a1.low, a1.high, a1.sum);
        // printf("a2.low=%d, a2.high=%d, a2.sum=%d\n", a2.low, a2.high, a2.sum);
        // printf("a3.low=%d, a3.high=%d, a3.sum=%d\n", a3.low, a3.high, a3.sum);
        if (a1.sum >= a2.sum && a1.sum >= a3.sum) {
            a->low = a1.low;
            a->high = a1.high;
            a->sum = a1.sum;
        }
        else if (a2.sum >= a1.sum && a2.sum >= a3.sum) {
            a->low = a2.low;
            a->high = a2.high;
            a->sum = a2.sum;
        }
        else {
            a->low = a3.low;
            a->high = a3.high;
            a->sum = a3.sum;
        }
        //printf("a->low=%d, a->high=%d, a->sum=%d\n", a->low, a->high, a->sum);
    }
}
void FindMaxCrosSub(int A[], int low, int mid, int high, struct ans *a)
{
    int leftsum = A[mid], rightsum = A[mid + 1];
    int minleft, i, maxright, j;
    int sum = 0;
    minleft = i = mid;
    maxright = j = mid + 1;
    while (i >= low) {
        sum += A[i];
        if (sum>leftsum) {
            leftsum = sum;
            minleft = i;
        }
        i--;
    }
    sum = 0;
    while (j <= high) {
        sum += A[j];
        if (sum>rightsum) {
            rightsum = sum;
            maxright = j;
        }
        j++;
    }
    a->low = minleft;
    a->high = maxright;
    a->sum = (leftsum + rightsum);
}