AtCoder Regular Contest 060--D - 桁和 / Digit Sum--数论+思维

D - 桁和 / Digit Sum

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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
• 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)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.

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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.

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Sample Input 1 Copy

Copy

87654
30

Sample Output 1 Copy

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10

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Sample Input 2 Copy

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87654
138

Sample Output 2 Copy

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100

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Sample Input 3 Copy

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87654
45678

Sample Output 3 Copy

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-1

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Sample Input 4 Copy

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31415926535
1

Sample Output 4 Copy

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31415926535

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Sample Input 5 Copy

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1
31415926535

Sample Output 5 Copy

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-1

给出n和s求一个进制b使得n在b进制下各位数之和为s

详细思路：

第一种情况
    x0+x1*b^1+x2*b^2+...=n
    x0+x1+x2+...=s
    这个时候b^2<=n
    所以b<=根号n
    这个时候暴力枚举即可

第三种情况暴力枚举
    x0+x1*b^1=n;
    x0+x1=s;
    这个时候联立二式子
    得到x1*(b-1)=n-s
    我们枚举x1，然后判断是否成立，包括题目条件以及进制常识

 

#include
using namespace std;
#define ll long long
ll n,s;
bool check(ll base)
{
    ll tmp=n;
    ll ans=0;
    while(tmp)
    {
        ans+=tmp%base;
        tmp/=base;
    }
    return ans==s;
}
bool check(ll i,ll sum)
{
    ll a1=s-i;//x0
    ll a2=sum/i+1;//b
    if(a1>=0&&a1=1;i--)
    {
        if(!(sum%i)&&(check(i,sum)))
        {
            printf("%lld\n",sum/i+1);
            return 0;
        }
    }
    printf("-1\n");
    return 0;
}