Description
Today is Yukari's n-th birthday. Ran and Chen hold a celebration party for her. Now comes the most important part, birthday cake! But it's a big challenge for them to place n candles on the top of the cake. As Yukari has lived for such a long long time. Though she herself insists that she is a 17-year-old girl.
To make the birthday cake look more beautiful, Ran and Chen decide to place them like r ≥ 1 concentric circles. They place ki candles equidistantly on the i-th circle, where k ≥ 2, 1 ≤ i ≤ r. And it's optional to place at most one candle at the center of the cake. In case that there are a lot of different pairs of r and k satisfying these restrictions, they want to minimize r × k. If there is still a tie, minimize r.
Input
There are about 10,000 test cases. Process to the end of file.
Each test consists of only an integer 18 ≤ n ≤ 1012.
Output
For each test case, output r and k.
Sample Input
18 111 1111
Sample Output
1 17 2 10 3 10
枚举+二分
刚开始还以为是线性规划……到网上搜了下才知道是枚举+二分
/* pow(int, double) 用于long long 的时候可能会出错,比如爆掉。 所以要重新写一个计算long long 的pow函数 */ #include <iostream> #include <cmath> using namespace std; long long powForLL(long long a, int b) //long long 的pow函数 { long long ans = 1; for (int i = 0; i < b; ++i) ans *= a; return ans; } int main() { long long n, r, k; while (cin >> n) { r = 1, k = n-1; for (int i = 2; i <= 45; ++i) //why 45? { long long ll = 2, rr = pow(n, 1.0/i), mm; // ll rr mm都是关于k的 i是关于r的 while (ll <= rr) { mm = (long long)(ll+rr)/2; long long ans = (mm-powForLL(mm,(double)i+1))/(1-mm);//等比数列求和公式别说你不懂 if (ans == n || ans == n-1) //中间不插蜡烛为n,插就n-1 { if (mm*i < r*k) { r = i, k = mm; } break; } else if (ans > n) { rr = mm-1; } else { ll = mm+1; } } } cout << r << ' ' << k << endl; } }