hdu 4430 Yukari's Birthday 枚举+二分

注意会超long long

开i次根号方法，te=(ll)pow(n,1.0/i);

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3262    Accepted Submission(s): 695

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.

 

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

 

Source

2012 Asia ChangChun Regional Contest

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #include<string>
 #include<iostream>
 #define maxi(a,b) (a)>(b)?(a):(b)
 #define mini(a,b) (a)<(b)?(a):(b)
 #define N 1000005
 #define mod 10000
 #define ll long long

 using namespace std;

 ll n;
 ll ans;
 ll r,k;
 ll rr,kk;

 void ini()
 {
     ans=n-;
     r=;
     k=n-;
 }

 void cal3(ll x)
 {
     ll te,d;
     te=+*x;
     d=sqrt(te);
     if(d*d==te){
         kk=(d-);
         if(kk%==){
             kk/=;
             if(kk*<ans){
                 ans=kk*;
                 k=kk;r=;
             }
         }
     }
 }

 ll cal(ll a,ll cnt)
 {
     ll re=;
     while(cnt)
     {
         if(cnt&){
             re=(re*a);
             cnt--;
         }
         cnt/=;
         a=a*a;
     }
     return re;
 }

 int cal2(ll a,ll en,ll now)
 {
     ll i;
     for(i=en;i>=;i--){
         now+=cal(a,i);
         if(now>n) return ;
     }
     if(now<n-) return ;
     if(now==n- || now==n){
         if(a*(en+)<ans){
             ans=a*(en+);
             r=en+;
             k=a;
         }
         else{
             if(a*(en+)==ans && (en+)<r){
                 r=en+;
                 k=a;
             }
         }
     }
     return ;
 }

 void solve()
 {
     ll te,temp;
     cal3(n);
     cal3(n-);
     ll i;
     for(i=;i<=;i++){
         //temp=cal(2ll,i);
         //if(temp>n) break;
         te=(ll)pow(n,1.0/i);
         for(kk=te;kk>=;kk--){
             temp=cal(kk,i);
             if(temp>n) continue;
             if(cal2(kk,i-,temp)==){
                 break;
             }
         }
     }
 }

 void out()
 {
     printf("%I64d %I64d\n",r,k);
 }

 int main()
 {
     //freopen("data.in","r",stdin);
     //scanf("%d",&T);
    // while(T--)
     while(scanf("%I64d",&n)!=EOF)
     {
         ini();
         solve();
         out();
     }
     return ;
 }