Number of non-printable items

Given expression [X/2] + Y + x * y, where X and Y are positive integers. The brackets indicate the following integers, for example, [3/2] = 1, [5/2] = 2. Some positive integers can be expressed using the above expression, for example, positive integer 2. When x = y = 1 is obtained, 2 can be expressed (explained below: When x = y = 1, [X/2] + Y + x * Y = [1/2] + 1 + 1*1 =
0 + 1 + 1 = 2); some numbers can be expressed in multiple ways. For example, 13 can be expressed by X = 2 y = 4 and x = 3 Y = 3; some numbers cannot be expressed using this expression, such as 3. The Nth number starting from 1 cannot be expressed by this expression. We call it an. For example, a1 = 1 A2 = 3. Given N, evaluate. Input: N value 1 <= n <= 40 output: An % 1000000007 result (because the result is large, output an % 1000000007 result ). Function header C/C ++: int givean (int n); Java: public class main
{Public static int givean (int n ){}}

#include <iostream>#include <vector>using namespace std;class Node{    public:        Node(int a, int b)        {            y = b;            value = (a>>1) + b + a*b;        }        void increase(int x)        {            y++;            value = (x>>1) + y + x*y;        }    int y;    long long int value;};int givean(int n){    vector<Node> buf;    buf.push_back(Node(0,0));    buf.push_back(Node(1,2));    buf.push_back(Node(2,1));    long long int preNum = 2;    long long int curNum = 2;    int preCount = 1;    int curCount = 1;    int result = 1;    while (curCount < n)    {        int x = 1;        int y = buf[1].y;        int curNum = buf[1].value;                int size = buf.size();        int i = 2;        for (; i < size; i++)        {            if ((buf[i].value) == preNum)            {                buf[i].increase(i);            }            if ((buf[i].value) < curNum)            {                x = i;                y = buf[i].y;                curNum = buf[i].value;            }        }        if (i == (size-1))        {            buf.push_back(Node(size,1));        }        buf[x].increase(x);                if (curNum != preNum+1)        {            curCount = preCount + (curNum-preNum-1);            if (curCount >= n)            {                result = preNum + (n-preCount);                break;            }            else            {                preCount = curCount;            }        }        preNum = curNum;    }    return result;}int main(){    for (int i = 1; i <= 40; i++)    {        cout << givean(i) << endl;    }    return 0;}