AtCoder Regular Contest 060（D - 桁和 / Digit Sum，数学规律）

最新推荐文章于 2025-03-03 14:12:54 发布

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最新推荐文章于 2025-03-03 14:12:54 发布

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分类专栏：思维题泛做

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博客围绕整数函数f(b,n)展开，给出题目描述、约束条件、输入输出示例等。核心题意指出暴力求解会超时，可采用分小于Sqrt（N）和大于等于Sqrt（N）的技巧，根据表达式枚举b的可能值，还给出他人理解的参考链接。

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题目描述：

D - 桁和 / Digit Sum

Time Limit: 2 sec / Memory Limit: 256 MB

Score : 500500 points

Problem Statement

For integers b(b≥2)b(b≥2) and n(n≥1)n(n≥1), let the function f(b,n)f(b,n)

be defined as follows:

f(b,n)=nf(b,n)=n, when n<bn<b
f(b,n)=f(b,floor(n/b))+(n mod b)f(b,n)=f(b,floor(n/b))+(n mod b), when n≥bn≥b

Here, floor(n/b)floor(n/b) denotes the largest integer not exceeding n/bn/b, and n

mod bn mod b denotes the remainder of nn divided by bb.

Less formally, f(b,n)f(b,n) is equal to the sum of the digits of nn written in base bb.

For example, the following hold:

f(10,87654)=8+7+6+5+4=30f(10,87654)=8+7+6+5+4=30
f(100,87654)=8+76+54=138f(100,87654)=8+76+54=138

You are given integers nn and ss. Determine if there exists an integer b(b≥2)

such that f(b,n)=sf(b,n)=s. If the answer is positive, also find the smallest such bb.

Constraints

1≤n≤10111≤n≤1011
1≤s≤10111≤s≤1011
n,sn,s are integers.

Input

The input is given from Standard Input in the following format:
nn
ss
Output

If there exists an integer b(b≥2)b(b≥2) such that f(b,n)=sf(b,n)=s, print the

smallest such bb. If such bb does not exist, print -1 instead.

Sample Input 1 Copy

Copy
87654
30
Sample Output 1 Copy

Copy
10
Sample Input 2 Copy

Copy
87654
138
Sample Output 2 Copy

Copy
100
Sample Input 3 Copy

Copy
87654
45678
Sample Output 3 Copy

Copy
-1
Sample Input 4 Copy

Copy
31415926535
1
Sample Output 4 Copy

Copy
31415926535
Sample Input 5 Copy

Copy
1
31415926535
Sample Output 5 Copy

Copy
-1

核心题意：

这题目一上来肯定先想暴力，但很明显会超时。然后这里好像有一个比较常用的技巧，就是分小于Sqrt（N）和大于等于Sqrt（N）考虑。然后根据题目给出的表达式枚举b的可能值。

其他人的理解：

https://blog.youkuaiyun.com/forever_shi/article/details/83278036

https://www.cnblogs.com/zjp-shadow/p/9851787.html

代码实现：

#include<bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
#define io ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
const int N=2e5+100;
using namespace std;
int judge(LL tmp,LL n,LL s)
{
	LL sum=0;
	while(n)
	{
		sum+=n%tmp;
		if(sum>s)return 0;
		n/=tmp;
	}
	if(sum==s)return 1;
	return 0;
}
int  main()
{
	io;
	LL n,s;
	while(cin>>n>>s)
	{
		if(n==s)
		{
			cout<<n+1<<endl;
			continue;
		}
		else if(s>n)
		{
			cout<<-1<<endl;
			continue;
		}
		LL f=0;
		LL up=sqrt(n)+1;
		for(LL i=2;i<=up;i++)
		{
			if(judge(i,n,s))
			{
				f=i;
				break;
			}
		}
		if(f)
		{
			cout<<f<<endl;
			continue;
		}
		else
		{
			LL up=sqrt(n-s);
			LL ans=999999999999999999;
			for(LL i=1;i<=up;i++)
			{
				
				if((n-s)%i!=0)continue;
				LL tmp=i+1;
				if(judge(tmp,n,s)&&((n-s)/i)==(n/tmp))
				{
					
					f=tmp;
					ans=min(ans,f);
				}
				else if(judge((n-s)/i+1,n,s)&&(i==n/((n-s)/i+1)))
				{
					
					f=(n-s)/i+1;
					ans=min(ans,f);
				}
			}
			if(f)cout<<ans<<endl;
			else cout<<-1<<endl;
		}
	}
	return 0;
}

The end；