HDU 4430 Yukari's Birthday 【二分查找】

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5676    Accepted Submission(s): 1370

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 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.

 

Input

There are about 10,000 test cases. Process to the end of file.
Each test consists of only an integer 18 ≤ n ≤ 10¹².

 

Output

For each test case, output r and k.

 

Sample Input

18 111 1111

 

Sample Output

1 17 2 10 3 10

 

Source

2012 Asia ChangChun Regional Contest

这个题有个比较坑的地放就是二分太大的时候会爆long long需要控制好上界

可以用ll high = pow(n, (double)1.0 / r);来紧上界防止爆long long

也可以在过程中判断long long是否溢出来解决

第一种做法：

#include<iostream>    
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<bitset>
#include<numeric>
#include<vector>
#include<string>
#include<iterator>
#include<cstring>
#include<functional>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;

const int maxn = 10010;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);

typedef pair<int, int> P;
typedef long long ll;
typedef unsigned long long ull;


int main()
{
    ll n;
    while (~scanf("%lld", &n))
    {
        ll ans = n - 1, ans1 = 1, ans2 = n - 1;
        for (int r = 2; r <= 41; r++)
        {
            ll low = 1;
            ll high = pow(n, (double)1.0 / r);
            ll mid;
            while (low <= high)
            {
                mid = (low + high) >> 1;
                ll tmp = 1, sum = 0;
                for (int i = 1; i <= r; i++)
                {
                    tmp *= mid;
                    sum += tmp;
                }
                if (sum == n || sum == (n - 1))
                {
                    if (mid*r < ans)
                    {
                        ans = mid*r;
                        ans1 = r;
                        ans2 = mid;
                    }
                }
                if (sum > n)
                {
                    high = mid - 1;
                }
                else
                {
                    low = mid + 1;
                }
            }
        }
        printf("%lld %lld
", ans1, ans2);
    }
}

第二种做法：

#include<iostream>    
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<bitset>
#include<numeric>
#include<vector>
#include<string>
#include<iterator>
#include<cstring>
#include<functional>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;

const int maxn = 10010;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);

typedef pair<int, int> P;
typedef long long ll;
typedef unsigned long long ull;


int main()
{
    ll n;
    while (~scanf("%lld", &n))
    {
        ll ans = n - 1, ans1 = 1, ans2 = n - 1;
        for (int r = 2; r <= 41; r++)
        {
            ll low = 1;
            ll high = n - 1;
            ll mid;
            while (low <= high)
            {
                mid = (low + high) >> 1;
                ll tmp = 1, sum = 0;
                bool flag = 0;
                for (int i = 1; i <= r; i++)
                {
                    if (tmp > (n / mid))
                    {
                        flag = 1;
                        break;
                    }
                    tmp *= mid;
                    sum += tmp;
                }
                if (!flag&&(sum == n || sum == (n - 1)))
                {
                    if (mid*r < ans)
                    {
                        ans = mid*r;
                        ans1 = r;
                        ans2 = mid;
                    }
                }
                if (flag||sum > n)
                {
                    high = mid - 1;
                }
                else
                {
                    low = mid + 1;
                }
            }
        }
        printf("%lld %lld
", ans1, ans2);
    }
}