HDU 4430 Yukari's birthday enumeration + binary

It will be too long.

Open the I root number method, 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 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 input18 111 1111 sample output1 17 2 10 3 10 source2012 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 }

 

HDU 4430 Yukari's birthday enumeration + binary