ARC060 Digit Sum II

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Problem Statement

For integers b(_b_≥2) and n(_n_≥1), let the function f(b,n) be defined as follows:

• f(b,n)=n, when n<b
• f(b,n)=f(b, floor(n_⁄_b))+(n mod b), when n_≥_b

Here, floor(n_⁄_b) denotes the largest integer not exceeding n_⁄_b, and n mod b denotes the remainder of n divided by b.

Less formally, f(b,n) is equal to the sum of the digits of n written in base b. For example, the following hold:

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

You are given integers n and s. Determine if there exists an integer b(_b_≥2) such that f(b,n)=s. If the answer is positive, also find the smallest such b.

Constraints

• 1≤_n_≤10¹¹
• 1≤_s_≤10¹¹
• n, s are integers.

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Input

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

n
s

Output

If there exists an integer b(_b_≥2) such that f(b,n)=s, print the smallest such b. If such b does not exist, print -1 instead.

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Sample Input 1

87654
30

Sample Output 1

10

题意

给定一个(n)和一个(s),找到一个最小的(b)使得(n)在(b)进制下各位之和为(s).

分析

假设(n)表示为(b)进制之后各位上的数字从低位到高位分别为(a_0,a_1,a_2,...),显然我们可以得到如下两个等式

[a_0+a_1b+a_2b^2+a_3b^3+...=n ag{1} ]

[a_0+a_1+a_2+a_3+... =s ag{2} ]

并且对于(forall i in [0,+infty])都有(a_i<b igwedge a_i>=0);

• 当(1)式中的最高次幂至少二次时,(bleq sqrt{n}),所以我们可以枚举(b)的值从(2)到(sqrt n),每次求出(b)进制下各位数字之和看是否等于(s)
• 当(1)式中的最高次幂为1次时,有$$
  left{
  egin{aligned}
  a_0+a_1b=n
  a_0+a_1=s
  end{aligned}
  ight.

[化简后有$a_1(b-1)=n-s$,所以$a_1$和$b-1$均为$n-s$的因数.所以我们可以枚举所有$n-s$的因数,并取出所有满足条件的$b$的最小值. + 当最高次幂为0次时,必须要满足$n=s$. ### 代码 ```cpp #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> #include <cmath> using namespace std; typedef long long ll; ll n,s; int main(int argc, char const *argv[]) { scanf("%lld%lld", &n,&s); ll b=1; for (b = 2; b*b <= n; ++b) { ll t=n; ll sum=0; while(t){ sum+=t%b; t/=b; } if(sum==s){ printf("%lld ", b); return 0; } } if(n<s){ printf("-1 "); return 0; } else if(n==s){ printf("%lld ", n+1); return 0; } ll i=1; for (i = 1; i*i <= n-s; ++i) { if((n-s)%i==0){ ll b=i+1; ll a1=(n-s)/i,a0=s-a1; if(a1<b&&a0<b&&a0>=0){ printf("%lld ", b); return 0; } } } for ( ; i >= 1; --i) { if((n-s)%i==0){ ll j=(n-s)/i; ll b=j+1; ll a1=(n-s)/j,a0=s-a1; if(a1<b&&a0<b&&a0>=0){ printf("%lld ", b); return 0; } } } printf("-1 "); return 0; } ```]