The high accuracy of C ++ is too painful ....
Yukari's birthday
Time Limit: 12000/6000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 2891 accepted submission (s): 604
Problem descriptiontoday 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 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.
Inputthere are about 10,000 test cases. process to the end of file.
Each test consists of only an integer 18 ≤ n ≤ 1012.
Outputfor each test case, output R and K.
Sample Input
181111111
Sample output
1 172 103 10
Source2012 Asia Changchun Regional Contest
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long int LL;LL n;LL check(LL r){ LL low=2,high=n,mid,ret=-1; while(low<=high) { mid=(low+high)/2; if(((LL)(pow(mid*1.,r-1.))>n||(LL)(pow(mid*1.,r-1.))<0) ||(LL)(pow(mid*1.,r*1.))>n||(LL)(pow(mid*1.,r*1.))<0) { high=mid-1; continue; } LL temp=mid*(1-(LL)(pow(mid*1.,r*1.)))/(1-mid); if(temp<=n-1||temp<=n) { if(temp==n||temp==n-1) ret=mid; low=mid+1; } else { high=mid-1; } } return ret;}int main(){while(scanf("%I64d",&n)!=EOF){ LL R=1,K=n-1,ans=(1LL<<60); for(LL r=2;r<40;r++) { LL k=check(r); if(k==-1) continue; if(ans>r*k) { R=r; K=k; } } printf("%I64d %I64d\n",R,K);} return 0;}
Hdoj 4430 Yukari's birthday