Array seeking the maximum value and the second largest

A seek array [] and the second largest maximum value in the interval [lo, hi) of.
Returns the index of the maximum index and the second largest value x1 x2

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Iteration 1:
(1 iterative codes are bug, if A [lo] as the maximum value, the result x1, x2 are LO)

void max2(int A[], int lo, int hi, int &x1, int &x2)
{
for(x1=lo, int i=lo+1; i<hi; i++)
if( A[i] > A[x1])
x1=i;
for(x2=lo, int i=lo+1; i<x1; i++)
if( A[i] > A[x2])
x2=i;
for(int i=x1+1; i<hi; i++)
if( A[i] > A[x2])
x2=i;
}

Iteration 2:
(1 iterations will not improve the relative complexity of the worst-case time)

void swap(int &x, int &y)
{ int t=x; x=y; y=t;}

void max2(int A[], int lo, int hi, int &x1, int &x2)
{
	if(A[x1=lo] < A[x2=lo+1]) //把x1设为最大值下标，x2设为次大值下标
		swap(x1,x2);
		
	for(int i=lo+2; i<hi; i++)
		if(A[i]>A[x2])  //先与A[x2]比较
			if(A[x2=i]>A[x1]) //若比A[x2]大，再把令x2指向i，然后A[x2]与A[x1]比较
				swap(x1,x2); //若比A[x1]也大，则交换x1，x2， 保持x1为最大，x2为次大
}

Recursion:
(divide and conquer)
(the array into left and right two arrays, each seeking about two times the maximum value and a large value of the array, then selecting the maximum value and the second largest value from the result)

void swap(int &x, int &y)
{
	int t = x; x = y; y = t;
}

void max2(int A[], int lo, int hi, int &x1, int &x2)
{
	if (lo + 2 == hi)//区间内只剩2个元素，为递归基之一
	{
		if (A[x1 = lo] < A[x2 = lo + 1])
			swap(x1, x2);
		return;
	}
	if (lo + 3 == hi)//区间只剩3个元素，为递归基之一
	{
		if (A[x1 = lo] < A[x2 = lo + 1])
			swap(x1, x2);
		if (A[lo + 2] > A[x2])
			if (A[x2 = lo + 2] > A[x1])
				swap(x1, x2);
		return;
	}

	//分而治之
	int mi = (lo + hi) / 2;
	int x1l, x2l, x1r, x2r;
	max2(A, lo, mi, x1l, x2l);
	max2(A, mi, hi, x1r, x2r);
	if (A[x1l]>A[x1r])
	{
		x1 = x1l;
		x2 = A[x1r]>A[x2l] ? x1r : x2l;
	}
	else
	{
		x1 = x1r;
		x2 = A[x2r]>A[x1l] ? x2r : x1l;
	}
}

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Test code:

#include <stdio.h>
void max2(int A[], int lo, int hi, int &x1, int &x2);

int main()
{
	int arr[100];
	int n;
	while (scanf("%d", &n) == 1)
	{
		for (int i = 0; i < n; i++)
			scanf("%d", arr + i);

		printf("\n数组是：\n");
		for (int i = 0; i < n; i++)
			printf("%d ", arr[i]);
		int x1, x2;
		max2(arr, 0, n, x1, x2);
		printf("\n最大值：%d  次大值：%d\n", arr[x1], arr[x2]);
	}
	return 0;
}

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ps: mi Why can this calculation mi = (LO + hi) / 2
(although this problem is very simple, but I had tangled for a long time, they still write it down.)
The most intuitive algorithm mi = lo + (hi-lo ) / 2
simplifying
mi The = LO + (Hi-LO) / 2
= LO + Hi / 2 - LO / 2
= (2 * LO) / 2 + Hi / 2 - LO / 2
= (2 * LO + Hi - LO ) / 2
= (LO + Hi) / 2
may be intuitively understood by the following image