这周讲的题目是二分，我就简单的说一下二分，二分就是在解空间有序的前提下，对半逐步逼近答案
这种方法很实在，很好用

Yukari’s Birthday
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 k i 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 ≤ 10 12.
Output
For each test case, output r and k.
Sample Input

18
111
1111

Sample Output

1 17
2 10
3 10

这个思路就是确定r (k>=2 ,所以r<=40）再二分查找k…
注意数据比较大时候容易溢出，神奇的是：cout好像不能用

#include <iostream>
#include <cstring>
#include <math.h>
#include <cstdio>
#define inf 0x3f3f3f3f

using namespace std;
long long  n;
int check(int k,int r){
    long long e=1,i,sum=0;
    for(i=1;i<=r;i++){
        e=e*k;
        sum+=e;
    }
    if(sum+1<n)return 0;
    if(sum>n)return -1;
    if(sum==n||sum+1==n)return 1;
}
int main()
{
    while(cin>>n){
    long long r,st,ed,mid,res=n-1,k=n-1,t=1;
    for(r=2;r<=60;r++){
        st=0;
        ed=pow(1e13,1.0/r)+1;
        while(ed-st>1){
        mid=st+(ed-st)/2;
        if(check(mid,r)==-1) ed=mid;
        if(check(mid,r)==0)st=mid;
        if(check(mid,r)==1) {
            if(mid*r<res){
             k=mid ;
             t=r;
            }
            if(mid*r==res){
                if(r<t){
                    t=r;
                    k=mid;
                }
            }
        break;
        }
        }
    }
    printf("%lld %lld\n",t,k);
    }

    return 0;
}