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

181111111

 

Sample Output

1 172 103 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\n", 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\n", ans1, ans2);    }}