滴滴编程题解C++

1.先求异或的“累异或”矩阵，类似累加，从i开始向后，一直找到异或成0的坐标。然后贪心求不重复区间。C+代码如下：

#include <iostream>
using namespace std;
int main()
{
	int n = 0;
	cin >> n;
	int *num = new int[n];
	int *num2 = new int[n];
	for (int i = 0; i < n; ++i)
		num2[i] = -1;
	for (int i = 0; i < n; ++i)
	{
		cin >> num[i];
		if (num[i] == 0)
		{
			num2[i] = i;
			continue;
		}
		for (int j = 0; j < i; ++j)
		{
			if (num2[j] == -1)
			{
				num[j] ^= num[i];
				if (num[j] == 0)
					num2[j] = i;
			}
		}
	}
	int output = 0;
	int k = 0;
	while (k < n)
	{
		int min = n;
		for (int i = k; i < n; ++i)
		{
			if (num2[i] != -1 && num2[i] < min)
			{
				min = num2[i];
			}
		}
		if(min == n)
			break;
		++output;
		k = min + 1;
	}
	cout << output;
	return 0;
}

2.剑指offer原题

#include <iostream>
using namespace std;
int getMin(int num1, int num2, int num3)
{
	int min = num1 < num2 ? num1 : num2;
	min = min < num3 ? min : num3;
	return min;
}
int main()
{
	int N = 0;
	cin >> N;
	if (N <= 0)
	{
		cout << 0;
		return 0;
	}
	int *num = new int[N];
	num[0] = 1;
	int index = 1;
	int *num2 = num;
	int *num3 = num;
	int *num5 = num;
	while (index < N)
	{
		int min = getMin(*num2 * 2, *num3 * 3, *num5 * 5);
		num[index] = min;
		while (*num2 * 2 <= min)
			++num2;
		while (*num3 * 3 <= min)
			++num3;
		while (*num5 * 5 <= min)
			++num5;
		++index;
	}
	cout << num[N - 1];
	return 0;
}