HDOJ 4430 Yukari & #39; s Birthday, hdojyukari

HDOJ 4430 Yukari's Birthday, hdojyukari

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;}

Which one can help me check what is wrong with HDOJ1002?

Let's show you some test data, right? 1. First, consider how many digits can you add to the question: 1000 digits 2.000 00 = 03.01 0 = 14.199 01 = 2005.990 110 1100 =

Why is hdoj1002 always Presentation Error? The format is correct many times. It's the same. I don't know where the Error is.

You read the question incorrectly. The question says "Output a blank line between two test cases. "It means there is a blank line between the two sets of data output. Your code does have a blank line after each set of output data, which is why PE. After knowing this, you can easily solve it yourself. In addition, it is recommended that you take a look at this document "High Quality C/C ++ programming". I believe it will be helpful for your code writing. Clear code writing is equally important to you.