TOJ 4373 HDU 4430 ZOJ 3665 Yukari ' s Birthday/two min

Yukari ' s Birthday time limit (normal/java): 1000ms/3000ms Run memory limit: 65536KByte

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 Placen candles on the top of the cake. As Yukari have lived for such a long long time. Though she herself insists that she's a 17-year-old girl.

The birthday cake look more beautiful, Ran and Chen decide to place them liker≥1 concentric circles. They place Ki candles equidistantly on thei-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 is a lot of different pairs ofr and K satisfying these restrictions, they want to minimizerxk. If There is still a tie, minimize R.

input

There is about the test cases. Process to the end of file.

Each test consists is an integer 18≤n≤1012.

Output

For each test case, output R and K.

Sample Input

1111 111

Sample Output

1
2
3 10

Assuming k minimum is 2 then R Max 45 2 45 times more than 10 times

Then enumerate r binary K

The two-time lvalue L is obviously from 2 R words I start from 100w because 100w squared is greater than or equal to n maximum

For the left and right range of the dichotomy is not very well judged

Rotten Code

 #include <stdio.h> const long Long MAX = 1000010; long long N;
	A long long check (long long K,long long R) {Long Long res = 0,i,m = k;
		for (i = 1;i <= r; i++) {res + = m;
		printf ("%i64d\n", res);
		M *= K;
	if (res > N) return max * MAX;
} return res;
	A long long erfen (long long L,long long R,long long k) {if (k = = 1) return n-1;
		while (L <= R) {Long Long M = (L + r) >> 1;
		Long Long res = check (m,k);
		printf ("%i64d\n", res);
		
		if (res = = N | | res = = n-1) return m;
		if (res > N) r = m-1;
	else L = m + 1;
} return-1;
	} int main () {Long long l,r,i,r,k,rr,kk;
		while (scanf ("%lld", &n)!=eof) {RR = MAX;
		KK = MAX;
			for (i = 1;i <=; i++) {L = 2;
			R = 1000000;
			Long Long res = Erfen (l,r,i);
					if (res! =-1) {if (res * i < RR * KK | | res * i = = RR * KK && i < RR) {rr = i;
				KK = res;
	}}} printf ("%lld%lld\n", RR,KK);
} return 0; }
	#include <stdio.h>
const long long MAX = 1000010;
long long n;

long long check (long long k, long long r)
{
long long res = 0, i, m = k;
for (i = 1; i <= r; i ++)
{
res + = m;
// printf ("% I64d \ n", res);
m * = k;
if (res> n)
return MAX * MAX;
}
return res;
}
long long erfen (long long l, long long r, long long k)
{
if (k == 1)
return n-1;
while (l <= r)
{
long long m = (l + r) >> 1;
long long res = check (m, k);
// printf ("% I64d \ n", res);
if (res == n || res == n-1)
return m;
Hell
if (res> n)
r = m-1;
else
l = m + 1;
}
return -1;
}
int main ()
{
long long l, r, i, R, K, RR, KK;
while (scanf ("% lld", & n)! = EOF)
{
RR = MAX;
KK = MAX;
for (i = 1; i <= 40; i ++)
{
l = 2;
r = 1000000;
long long res = erfen (l, r, i);
if (res! = -1)
{
if (res * i <RR * KK || res * i == RR * KK && i <RR)
{
RR = i;
KK = res;
}
}
}
printf ("% lld% lld \ n", RR, KK);
}
return 0;
}