题意：给定一个n（18 ≤ n ≤ 10^12），一个等比数列k + k^2 + .......+ k^r = n 或者 = n-1，求出最小的k*r，如果最小的不唯一，则取r更小的
分析：两个未知数，r,k，很明显，r的范围只有几十而已，所以枚举r；k的范围很大，需要二分...................
二分k的上界依情况而定 ： pow(n,1.0/i)；
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <climits>//形如INT_MAX一类的
#define MAX 100005
#define INF 0x7FFFFFFFFFFFFFFF
#define REP(i,s,t) for(int i=(s);i<=(t);++i)
#define ll __int64
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define L(x) x<<1
#define R(x) x<<1|1
# define eps 1e-5
//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂
using namespace std;
__int64 n; struct node {
__int64 k,ans;
int r;
}minn; __int64 pow2(__int64 mid,int r) {
__int64 t = mid;
for(int i=0; i<r-1; i++) {
mid *= t;
}
return mid;
} __int64 search(int r,__int64 tmp,__int64 high) {
__int64 low = 2,mid;
while(low <= high ) {
mid = (low + high) >> 1;
__int64 cal = (pow2(mid,(r+1)) - mid) / (mid-1);
//cout << "k: " << mid << " r: " << r << " cal: " << cal << endl;
if(cal > tmp) {
high = mid - 1;
} else if(cal < tmp) {
low = mid + 1;
} else {
return mid;
}
}
return -1;
} int main(){
while(cin >> n) {
int high = log(n) / log(2);
minn.ans = n-1;
minn.r = 1;
minn.k = n-1;
for(int i=2; i<=high; i++) {
int tmp = pow(n,1.0/i);
__int64 tmp1 = search(i,n,tmp);
__int64 tmp2 = search(i,n-1,tmp);
// cout << "i: " << i << " tmp: " << tmp << " tmp1: " << tmp1 << " tmp2: " << tmp2 << endl;
if(tmp1 != -1 && i * tmp1 <= minn.ans) {
if(tmp1 * i == minn.ans && i < minn.r) {
minn.r = i; minn.k = tmp1;
} else if(i * tmp1 < minn.ans){
minn.ans = i * tmp1; minn.r = i; minn.k = tmp1;
}
}
if(tmp2 != -1 && i * tmp2 <= minn.ans) {
if(tmp2 * i == minn.ans && i < minn.r) {
minn.r = i; minn.k = tmp2;
} else if(i * tmp2 < minn.ans){
minn.ans = i * tmp2; minn.r = i; minn.k = tmp2;
}
}
}
printf("%d %I64d\n",minn.r,minn.k);
}
return 0;
}