Binomial coefficients

B - Binomial coefficients

Time Limit:3000MS    Memory Limit:0KB    64bit IO Format:%lld & %llu

Submit Status Practice UVALive 5900

Description

Gunnar is quite an old and forgetful researcher. Right now he is writing a paper on security in social networks and it actually involves some combinatorics. He wrote a program for calculating binomial coefficients to help him check some of his calculations.

A binomial coefficient is a number

n = ,

where n and k are non-negative integers.

Gunnar used his program to calculate n and got a number m as a result. Unfortunately, since he is forgetful, he forgot the numbers n and k he used as input. These two numbers were a result of a long calculation and they are written on one of many papers lying on his desk. Instead of trying to search for the papers, he tried to reconstruct the numbers n, k from the output he got. Can you help him and find all possible candidates?

Input

On the first line a positive integer: the number of test cases, at most 100. After that per test case:

• one line with an integer m (2m10¹⁵): the output of Gunnar's program.

Output

Per test case:

• one line with an integer: the number of ways of expressing m as a binomial coefficient.
• one line with all pairs (n, k) that satisfy n = m. Order them in increasing order of n and, in case of a tie, order them in increasing order of k. Format them as in the sample output.

Sample Input

2
2
15

Sample Output

1
(2,1)
4
(6,2) (6,4) (15,1) (15,14)




主要学习别人的代码的STL的使用；


#include <stdio.h>
#include <vector>
#include <string.h>
#include <iostream>
#include <math.h>
#include <map>
#include <algorithm>

using namespace std;

typedef long long LL;
typedef pair<LL, LL> PLL;
#define MAX 1000000000000009LL

LL C[1000][1000];

void makeC()
{
    memset(C, 0, sizeof (C));
    for (int i = 0; i < 1000; i++)
    {
        C[i][0] = 1;
        for (int j = 1; j <= i; j++)
        {
            C[i][j] = C[i-1][j-1] + C[i-1][j];
            C[i][j] = min(C[i][j], MAX);
        }
    }
}

LL getC(LL a, LL b)
{
    LL ret = 1;
    for (LL i = 1; i <= b; i++)
    {
        LL tmp = ret * (a + 1 - i) / i;
        if (ret == tmp * i / (a + 1 - i))
        {
            ret = tmp;
        }
        else
        {
            return MAX;
        }
    }
    return ret;
}

vector<PLL> ans;
void gao(LL n)
{
    ans.clear();
    for (int i = 1; i < 1000; i++)
    {
        for (int j = 1; j <= i; j++)
        {
            if (C[i][j] == n)
            {
                ans.push_back(PLL(i, j));
            }
        }
    }
    for (int i = 1; i < 10; i++)
    {
        LL l = 1;
        LL r = n;
        while (l <= r)
        {
            LL m = (l + r) >> 1;
            LL t = getC(m, i);
            if (t == n)
            {
                ans.push_back(PLL(m,     i));
                ans.push_back(PLL(m, m - i));
                break;
            }
            else if (t > n)
            {
                r = m - 1;
            }
            else
            {
                l = m + 1;
            }
        }
    }
    sort(ans.begin(), ans.end());
    ans.resize(unique(ans.begin(), ans.end()) - ans.begin());
}

int main ()
{
    makeC();
    int T;
    cin >> T;
    while (T--)
    {
        LL n;
        cin >> n;
        gao(n);
        int sz = ans.size();
        cout << sz << endl;
        for (int i = 0; i < sz; i++)
        {
            if (i)
            {
                cout << " ";
            }
            cout << "(" << ans[i].first << "," << ans[i].second << ")";
        }
        cout << endl;
    }
    return 0;
}