Uva 1649 Binomial coefficients

有一定难度的一道题目，网上已经有大神给出了思路，其实也就是按照别人的方法得知最大的k值不会超过26，那么对于不同的k值，我们直接二分查找是否存在对应的n值即可，如果存在就保存对应的结果，如果不存在就进行下一轮的查找，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;typedef long long LL;int judge(LL mid,LL aim,LL k){LL r = 1;for (LL i = 1; i <= k; i++){if (aim*i / r / (mid - i + 1) == 0) return 1;//代表mid取大了r = r*(mid - i + 1) / i;}if (r == aim) return 0;//正好return -1;//mid取小了}int main(){int Case;cin >> Case;vector<pair<LL, LL>> res;while (Case--){res.clear();LL m;cin >> m;res.push_back(make_pair(m,1));res.push_back(make_pair(m,m-1));for (int i = 2; i <= 30; i++){LL left = i, right = m;while (left <= right){LL mid = (left + right) / 2;LL temp = judge(mid, m, i);if (temp == 0){res.push_back(make_pair(mid,i));res.push_back(make_pair(mid, mid - i));break;}else if (temp == 1){right = mid - 1;}else left = mid + 1;}}sort(res.begin(),res.end());res.resize(unique(res.begin(),res.end())-res.begin());cout << res.size() << endl;int amount = res.size() - 1;for (int i = 0; i < amount; i++){cout << "(" << res[i].first << "," << res[i].second << ") ";}cout << "(" << res[amount].first << "," << res[amount].second << ")\n";}return 0;}