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

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cmath>
  4 #include<cstring>
  5 #include<string>
  6 #include<iostream>
  7 #define maxi(a,b) (a)>(b)?(a):(b)
  8 #define mini(a,b) (a)<(b)?(a):(b)
  9 #define N 1000005
 10 #define mod 10000
 11 #define ll long long
 12 
 13 using namespace std;
 14 
 15 ll n;
 16 ll ans;
 17 ll r,k;
 18 ll rr,kk;
 19 
 20 void ini()
 21 {
 22     ans=n-1;
 23     r=1;
 24     k=n-1;
 25 }
 26 
 27 void cal3(ll x)
 28 {
 29     ll te,d;
 30     te=1+4*x;
 31     d=sqrt(te);
 32     if(d*d==te){
 33         kk=(d-1);
 34         if(kk%2==0){
 35             kk/=2;
 36             if(kk*2<ans){
 37                 ans=kk*2;
 38                 k=kk;r=2;
 39             }
 40         }
 41     }
 42 }
 43 
 44 ll cal(ll a,ll cnt)
 45 {
 46     ll re=1;
 47     while(cnt)
 48     {
 49         if(cnt&1){
 50             re=(re*a);
 51             cnt--;
 52         }
 53         cnt/=2;
 54         a=a*a;
 55     }
 56     return re;
 57 }
 58 
 59 int cal2(ll a,ll en,ll now)
 60 {
 61     ll i;
 62     for(i=en;i>=1;i--){
 63         now+=cal(a,i);
 64         if(now>n) return 0;
 65     }
 66     if(now<n-1) return 2;
 67     if(now==n-1 || now==n){
 68         if(a*(en+1)<ans){
 69             ans=a*(en+1);
 70             r=en+1;
 71             k=a;
 72         }
 73         else{
 74             if(a*(en+1)==ans && (en+1)<r){
 75                 r=en+1;
 76                 k=a;
 77             }
 78         }
 79     }
 80     return 2;
 81 }
 82 
 83 void solve()
 84 {
 85     ll te,temp;
 86     cal3(n);
 87     cal3(n-1);
 88     ll i;
 89     for(i=3;i<=45;i++){
 90         //temp=cal(2ll,i);
 91         //if(temp>n) break;
 92         te=(ll)pow(n,1.0/i);
 93         for(kk=te;kk>=2;kk--){
 94             temp=cal(kk,i);
 95             if(temp>n) continue;
 96             if(cal2(kk,i-1,temp)==2){
 97                 break;
 98             }
 99         }
100     }
101 }
102 
103 void out()
104 {
105     printf("%I64d %I64d\n",r,k);
106 }
107 
108 int main()
109 {
110     //freopen("data.in","r",stdin);
111     //scanf("%d",&T);
112    // while(T--)
113     while(scanf("%I64d",&n)!=EOF)
114     {
115         ini();
116         solve();
117         out();
118     }
119     return 0;
120 }

posted on 2014-10-17 20:12 njczy2010 阅读(220) 评论(0) 编辑收藏举报