題目連結

AcCode:

#include<bits/stdc++.h>
#define  M(a, b) make_pair(a, b)
using namespace std;
typedef long long LL;
typedef pair<long long , long long> pll;

LL m;
priority_queue<pll, vector<pll>, greater<pll> > q;

LL C(long long n, int k)
{
    int i;
    long long f = 1;
    for(i = 1; i <= k; i++){
        //一旦中間有哪一步超過了m那說明這個解就不存在了
        if (f /i > m / (n-i+1))  //這樣是對的
            return m+1;
        /*底下的都錯了：
        if(f > m)
        if(f*(n-i+1)/i > m)
        if (f * (n-i+1) > m * i)
           return m+1;
        */
        f *= (n-i+1);
        f /= i;
    }
    return f;
}

void solve()
{
    long long left, right, mid, t;
    int k;
    for(k = 1; C(2*k, k) <= m; k++){
        /*
        為什麼二分最開始的邊界是2*k 和 m
        答： C(k, n)函式的定義域是2*k ~ m ,在這一定義域內影象是單調遞減的。
        注意自變數是n, 不是k  !!!
        */
        left = k*2, right = m;
        while(left <= right){
            mid = (left + right) >> 1;
            t = C(mid, k);
            if(t == m){
                q.push(M(mid, k));
                if(mid == k*2) break;
                q.push(M(mid, mid-k));
                break;
            }
            else if(t < m)
                left = mid + 1;
            else
                right = mid - 1;
        }
    }

}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%lld", &m);
        solve();
        int len = q.size();
        if(len){
            cout << len << endl;
            for(int i = 0; i < len; i++){
                printf("(%lld,%lld)%c", q.top().first, q.top().second, i == len-1 ? '\n' : ' ');
                q.pop();
            }
        }
        else
            cout << "0" << endl;
    }
    return 0;
}