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;
}
