hdu 4430 Yukari's Birthday(二分+枚举)
Yukari's Birthday
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 883 Accepted Submission(s): 151
Problem 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.
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.
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
Source
Recommend
zhuyuanchen520
思路:本题数据是10^12用二分就行了,最多枚举不会超过64位,因为2^64>>10^12,因此枚举某个数的次数,看看是否有满足条件的r和k即可。
失误点:中心点可以放也可以不放。
#include<iostream> using namespace std; const long long oo=1000001; ///二分查找是否有满足条件的k long long bserch(int x,long long n) { if(x==1)return n; long long l=2,r=oo,mid,sum,bit; while(l<=r) { mid=(l+r)/2;sum=0;bit=1; for(int i=0;i<x;i++) {bit*=mid;sum+=bit; if(sum>n) { sum=oo*oo; break; } } if(sum==n)return mid; else if(sum>n)r=mid-1; else l=mid+1; } return 0; } int main() { long long r,k,ar,ak,n; while(cin>>n) { ar=ak=oo; for(int i=1;i<64;i++) { k=bserch(i,n); if(k!=0) { r=i; if(r*k<ar*ak) ar=r,ak=k; else if(r*k==ar*ak&&r<ar) ar=r,ak=k; } k=bserch(i,n-1); if(k!=0) { r=i; if(r*k<ar*ak) ar=r,ak=k; else if(r*k==ar*ak&&r<ar) ar=r,ak=k; } } cout<<ar<<" "<<ak<<"\n"; } }
The article write by nealgavin