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
Author: WU, Zejun
Contest: The 2012 ACM-ICPC Asia Changchun Regional Contest
长春现场赛时,第一道题就是做的这题。
就是枚举r,然后用二分来求对应的k.
由于题目上k>=2,所以r肯定不会很大,二分就可以了。
#include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> #include<math.h> using namespace std; long long solve(int r,long long sum) { long long left=2,right=(long long)sqrt((double)sum); while(left<=right) { long long mid=(left+right)/2; double mm=(pow(1.0*mid,r+1)-mid)/(mid-1); if(mm>((double)sum+1e-2)) { right=mid-1; continue; } long long temp=0; long long tt=1; for(int i=0;i<r;i++) { tt*=mid; temp+=tt; } if(temp==sum)return mid; else if(temp<sum)left=mid+1; else right=mid-1; } return 0; } int main() { long long n; long long ans,ansk; int ansr; while(scanf("%lld",&n)!=EOF) { ans=n-1; ansr=1; ansk=n-1; for(int r=2;r<=40;r++) { long long temp=solve(r,n); //printf("%d\n",temp); if(temp==0)continue; if(r*temp<ans) { ans=r*temp; ansr=r; ansk=temp; } else if(r*temp==ans&&r<ansr) { ansr=r; ansk=temp; } } for(int r=2;r<=40;r++) { long long temp=solve(r,n-1); if(temp==0)continue; if(r*temp<ans) { ans=r*temp; ansr=r; ansk=temp; } else if(r*temp==ans&&r<ansr) { ansr=r; ansk=temp; } } printf("%d %lld\n",ansr,ansk); } return 0; }