hdu 4430 枚举+二分范围运算失误

最新推荐文章于 2022-08-01 12:46:43 发布

AC_Lee

最新推荐文章于 2022-08-01 12:46:43 发布

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分类专栏：二分枚举

本文链接：https://blog.youkuaiyun.com/xiaoli_nu/article/details/73853344

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题目：

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5683 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

总共有N支蜡烛，要把它们放在蛋糕上蜡烛必须放成一个同心圆的形状，圆心可以选择放一个蜡烛或者不放从里到外第i层圆可以放k^i支蜡烛

问怎样放（选择K咯）可以使得r*k最小如果有多种方案，选择r最小的方案输出K和R

分析：

队友想到了用枚举+二分.....我们分析得k最小为2时，r最大也不会超过log(1e12)/log(2)=39 因此r的范围是1-39 而k的范围是2-1e12，所以应该枚举r,二分k，找出所有刚好能将蜡烛摆好的情况，记录最小值就可以了.....

代码：
二分写的太少了，实现花了好久.....

依然WA代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll inf=(ll)1<<63-1;//不减一是负数！！！！！！！！！
// 注意long long 输出格式，win测评系统用%I64d,linux 用%lld

int main(){//
    ll n;
    while(scanf("%lld",&n)==1){
        ll R,K,RK=inf;//RK初始化为(ll)1<<63爆炸
        ll t,ans;
        for(ll i=1;i<=64;++i){//枚举R
            ll l=2,r=1e6+30;//这里的r和之前的r重名...debug n遍 ..........
            ll mid;
            while(l<=r){//二分K
                mid=(l+r)>>1;
                t=1,ans=0;
                int f=0;
                for(ll j=1;j<=i;++j){
                    t*=mid;
                    ans+=t;
                    if(ans>n){f=1; break;}
                }
                if(f){//mid太大了
                    r=mid-1;
                }
                else if(ans==n){
                    //cout<<"i=:"<<i<<" mid=:"<<mid<<" i*mid:"<<i*mid<<" RK=:"<<RK<<endl;
                    if(i*mid<RK){ R=i; K=mid; RK=i*mid;}//之前没更新RK SB.....
                    else if((i*mid==RK&&i<R)){ R=i; K=mid; RK=i*mid;}//之前以为这种情况包含在上一种情况里面了...
                    break; //找到了=n的接下来也有可能找到=n-1的 而这里不break二分又不能处理，所以=n和=n-1的情况分开处理比较好
                }
                else l=mid+1;
            }
        }

        for(ll i=1;i<=64;++i){//枚举R
            ll l=2,r=1e6+30;//这里的r和之前的r重名...debug n遍 ..........
            ll mid;
            while(l<=r){//二分K
                mid=(l+r)>>1;
                t=1,ans=0;
                int f=0;
                for(ll j=1;j<=i;++j){
                    t*=mid;
                    ans+=t;
                    if(ans>n){f=1; break;}
                }
                if(f){//mid太大了
                    r=mid-1;
                }
                else if(ans==(n-1)){
                    //cout<<"i=:"<<i<<" mid=:"<<mid<<" i*mid:"<<i*mid<<" RK=:"<<RK<<endl;
                    if(i*mid<RK){ R=i; K=mid; RK=i*mid;}
                    else if((i*mid==RK&&i<R)){ R=i; K=mid; RK=i*mid;}
                    break;
                }
                else l=mid+1;
            }
        }
        printf("%lld %lld\n",R,K);
    }
	return 0;
}

因为参数r 以及i==1的情况没有特殊处理，疯狂WA

AC代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll inf=(ll)1<<63-1;//不减一是负数！！！！！！！！！
// 注意long long 输出格式，win测评系统用%I64d,linux 用%lld

int main(){//483MS	1688K
    ll n;
    while(scanf("%lld",&n)==1){
        ll R=1,K=n-1,RK=n-1;//RK初始化为(ll)1<<63爆炸
        ll t,ans;
        for(ll i=2;i<=40;++i){//枚举R
            ll l=2,r=1e6+30;//这里的r和之前的r重名...debug n遍 ..........
            //r=1e12+30  WA ?????????????????????????????????????
            ll mid;
            while(l<=r){//二分K
                mid=(l+r)>>1;
                t=1,ans=0;
                int f=0;
                for(ll j=1;j<=i;++j){
                    t*=mid;
                    ans+=t;
                    if(ans>n){f=1; break;}
                }
                if(f){//mid太大了
                    r=mid-1;
                }
                else if(ans==n){
                    //cout<<"i=:"<<i<<" mid=:"<<mid<<" i*mid:"<<i*mid<<" RK=:"<<RK<<endl;
                    if(i*mid<RK){ R=i; K=mid; RK=i*mid;}//之前没更新RK SB.....
                    else if((i*mid==RK&&i<R)){ R=i; K=mid; RK=i*mid;}//之前以为这种情况包含在上一种情况里面了...
                    break; //找到了=n的接下来也有可能找到=n-1的 而这里不break二分又不能处理，所以=n和=n-1的情况分开处理比较好
                }
                else l=mid+1;
            }
        }

        for(ll i=2;i<=40;++i){//枚举R
            ll l=2,r=1e6+30;//这里的r和之前的r重名...debug n遍 ..........
            ll mid;
            while(l<=r){//二分K
                mid=(l+r)>>1;
                t=1,ans=0;
                int f=0;
                for(ll j=1;j<=i;++j){
                    t*=mid;
                    ans+=t;
                    if(ans>n){f=1; break;}
                }
                if(f){//mid太大了
                    r=mid-1;
                }
                else if(ans==(n-1)){
                    //cout<<"i=:"<<i<<" mid=:"<<mid<<" i*mid:"<<i*mid<<" RK=:"<<RK<<endl;
                    if(i*mid<RK){ R=i; K=mid; RK=i*mid;}
                    else if((i*mid==RK&&i<R)){ R=i; K=mid; RK=i*mid;}
                    break;
                }
                else l=mid+1;
            }
        }
        cout<<R<<" "<<K<<endl;//两种方式都可以
        //printf("%lld %lld\n",R,K);
    }
	return 0;
}

队友1A代码，先贴过来：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Mod 1000000007
using namespace std;
typedef long long ll;
const double PI=acos(-1.0);
const double eps=1e-6;
const int INF=1000000000;
const int maxn=100;
ll n;
ll bi_search(int r, ll ans)
{
	ll left=2;
	ll right=ans;
	while(left<=right)
	{
		ll mid=(left+right)>>1;
		double t=(pow(mid*1.0,r+1)-mid)/(mid+1);
		if(t>((double)ans))
		{
			right=mid-1;
			continue;
		}
		ll temp=0;
		ll tt=1;
		for(int i=0;i<r;i++)
		{
			tt*=mid;
			temp+=tt;
		}
		if(temp==ans)return mid;
		else if(temp<ans)left=mid+1;
		else right=mid-1;
	}
	return 0;
}
int main()
{
	ll n;
	ll ans,k;
	int ar;
	while(scanf("%lld",&n)!=EOF)
	{
		ans=n-1;
		ar=1;
		k=n-1;
		for(int r=2;r<=40;r++)
		{
			ll t=bi_search(r,n);
			if(t==0)continue;
			if(r*t<ans)
			{
				ans=r*t;
				ar=r;
				k=t;
			}
			else if(r*t==ans&&r<ar)
			{
				ar=r;
				k=t;
			}
		}
		for(int r=2;r<=40;r++)
		{
			ll t=bi_search(r,n-1);
			if(t==0)continue;
			if(r*t<ans)
			{
				ans=r*t;
				ar=r;
				k=t;
			}
			else if(r*t==ans&&r<ar)
			{
				ar=r;
				k=t;
			}
		}
		printf("%d %lld\n",ar,k);
	}
	return 0;
}