Problem Link

Yukari’s Birthday
非常简单的二分题，给你一个数 n 让找 $r,k$ s,t$\sum_{i = 1}^rk^i ==n || n-1$，让你找满足条件的最小化$r*k$多解就找最小的$r$

分析

很简单的二分，枚举$r$，二分 k <script type="math/tex" id="MathJax-Element-16">k</script>
用快速冪的时候一定注意范围溢出，找了很久的一个bug。

AC code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include <set>
#define fi first
#define se second
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
LL n;
const int maxn  = 1e5+10;

PII ans;

LL power(LL x,LL n){
  LL res = 1;
  while (n) {
    if(n%2)res*=x;
    x*=x;
    n>>=1;
  }
  return res;
}


void solve(){
  LL ans_v = n-1;
  ans.fi = 1;ans.se = n-1;
  for(LL r = 2 ; r<=41 ; ++r){
    LL left = 2 ; LL right = (LL)pow(n,1.0/r);

    while (right>=left) {
        LL mid = (right+left)>>1;

        //std::cout << "power " << power(mid,r)<<endl;
        LL sum  = (mid-power(mid,r+1))/(1-mid);

        //std::cout << sum << '\n';
        if(sum == n || sum == n-1){
          //std::cout << r << " "<<mid<< endl;
          if(ans_v>r*mid){
            ans_v = r*mid;
            ans.fi = r;ans.se = mid;
          }
          break;
        }
        if(sum<n)left = mid+1;
        else right = mid-1;
    }
  }
}


int main()
{
    //freopen("in.txt","r",stdin);
   while (cin>>n) {
      solve();
      std::cout << ans.fi << " "<<ans.se<<endl;
      //printf("%I64d %I64d\n",ans.fi,ans.se );
    }

    return 0;
}