hdoj：2058

 

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;

struct node{
    int left;
    int right;
};
int main()
{
    long  N, M;
    long long sum = 0;
    vector<node> nodes;
    while (cin >> N >> M)
    {
        if (N == 0 && M == 0)
            break;

        for (long n = N; n >=1; n--)
        {
            if ((2*M-n*n+n)%(2*n)!=0)
                continue;
            long long a1 = (2 * M - n*n + n) / (2 * n);
            
            long long an = a1 + n - 1;
            if (a1 >= 1 && a1 <= N && an >= 1 && an <= N && a1 <= an)
            {
                //node node;
                //node.left = a1;
                //node.right = an;
                //nodes.push_back(node);
                cout << "[" << a1 << "," << an << "]" << endl;
            }
        }
        //for (node node : nodes)
        //{
        //    cout << "[" << node.left << "," << node.right << "]" << endl;
        //}
        cout << endl;
    }
    return 0;
}

 

超時

 

 

a1 = k

an = k + n - 1

M = n*(2k+n-1)/2

解的

k = M/n - (n-1)/2，這個不能這樣寫，要寫在一塊兒

(2M-n*n-n)/(2*n)。不然對於30/4 - 3/2，會忽略這樣的結果，若是先通分固然也就能夠。

 超時

 

利用a1>=1這個條件能夠減小循環次數

2M - n*n >n

2M > n*n - n>n*n

sqrt(2M)>n

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;

struct node{
    int left;
    int right;
};
int main()
{
    long  N, M;
    long long sum = 0;
    vector<node> nodes;
    while (cin >> N >> M)
    {
        if (N == 0 && M == 0)
            break;

        for (long n = sqrt(2*M); n >=1; n--)
        {
            if ((2*M-n*n+n)%(2*n)!=0)
                continue;
            long a1 = (2 * M - n*n + n) / (2 * n);
            
            long an = a1 + n - 1;
            if (a1 >= 1 && a1 <= N && an >= 1 && an <= N && a1 <= an)
            {
                //node node;
                //node.left = a1;
                //node.right = an;
                //nodes.push_back(node);
                cout << "[" << a1 << "," << an << "]" << endl;
            }
        }
        //for (node node : nodes)
        //{
        //    cout << "[" << node.left << "," << node.right << "]" << endl;
        //}
        cout << endl;
    }
    return 0;
}