HDU

题意：已知有n个蜡烛，过生日在蛋糕上摆蜡烛，将蜡烛围成同心圆，每圈个数为kⁱ，蛋糕中心最多可摆一个蜡烛，求圈数r和看，条件为r*k尽可能小的情况下，r尽可能小。

分析：n最大为10¹²，k最少为2，假设k为2，r最多为40，因此枚举r，二分k。

需要两个剪枝防止爆LL，

在计算ans=k¹+k²+……+k^r的过程中

(1)当k^r>n时，break，并向左区间继续搜索

(2))当ans>n时，break，并向左区间继续搜索

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
LL n;
struct Node{
    int r;
    LL k;
    bool operator < (const Node &rhs)const{
        return r * k < rhs.r * rhs.k || (r * k == rhs.r * rhs.k && r < rhs.r);
    }
}num[50];
LL deal(LL k, int r){
    LL ans = 0;
    LL tmp = 1;
    bool ok = true;
    for(int i = 1; i <= r; ++i){
        if(n / tmp < k){
            return -1;
        }
        tmp *= k;
        ans += tmp;
        if(ans > n){
            return -1;
        }
    }
    return ans;
}
LL solve(int r){
    LL L = 2, R = n;
    while(L <= R){
        LL mid = L + (R - L) / 2;
        LL tmp = deal(mid, r);
        if(tmp == n || tmp == n - 1) return mid;
        if(tmp == -1) R = mid - 1;
        else if(tmp < n - 1) L = mid + 1;
    }
    return -1;
}
int main(){
    while(scanf("%lld", &n) == 1){
        int cnt = 0;
        for(int r = 1; r <= 40; ++r){
            LL k = solve(r);
            if(k == -1) continue;
            num[cnt].r = r;
            num[cnt++].k = k;
        }
        sort(num, num + cnt);
        printf("%d %lld
", num[0].r, num[0].k);
    }
    return 0;
}