Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1993 Accepted Submission(s): 404

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 k

ⁱ 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 ≤ 10

¹².

Output

For each test case, output r and k.

Sample Input

18 111 1111

Sample Output

1 17 2 10 3 10

Source

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                 //ios_base::sync_with_stdio(false);//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define sz(v) ((int)(v).size())#define rep(i, a, b) for (int i = (a); i < (b); ++i)#define repf(i, a, b) for (int i = (a); i <= (b); ++i)#define repd(i, a, b) for (int i = (a); i >= (b); --i)#define clr(x) memset(x,0,sizeof(x))#define clrs( x , y ) memset(x,y,sizeof(x))#define out(x) printf(#x" %d\n", x)#define sqr(x) ((x) * (x))typedef long long LL;const LL INF = 100000000000000000;const double eps = 1e-8;const int maxn = 30000;int sgn(const double &x) { return (x > eps) - (x < -eps); }LL Pow(LL a,int b){ LL ans = 1; repf(i,1,b) ans*= a; return ans;}int main() { //freopen("in.txt","r",stdin); LL n; while(scanf("%I64d",&n) == 1) { LL k,r; LL Max = INF; LL ans1 = INF; LL ans2 = INF; repf(i,2,50) { r = i; LL L = 2; LL R = pow(n,1.0/i); k = -1; while(L <= R) { LL mid = (L + R)/2; LL temp = Pow(mid,i+1); if((temp - 1)%(mid - 1) == 0) { if((temp-1)/(mid-1) == n || (temp-1)/(mid-1) == n+1) { k = mid; break; } } if((temp-1)/(mid-1) <= n) L = mid + 1; if((temp-1)/(mid-1) >= n+1) R = mid - 1; } if(k == -1) { continue; } if(k*r < Max) { ans1 = r; ans2 = k; Max = k*r; } if(Max == k*r) { ans1 = min(ans1,r); } } if(ans1 == INF) { ans1 = 1; ans2 = n - 1; } cout<
                 
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