ARC-060D - Digit Sum - 数论

题解链接

https://www.lucien.ink/archives/230/

---

题目链接

https://arc060.contest.atcoder.jp/tasks/arc060_b

---

题目

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 f ( b , n ) = n , when n<b n < b
f(b,n)=f(b,floor(n⁄b))+(n mod b) f ( b , n ) = f ( b ,   f l o o r ( n ⁄ b ) ) + ( n   m o d   b ) , when n≥b n ≥ b
Here, floor(n⁄b) f l o o r ( n ⁄ b ) denotes the largest integer not exceeding n⁄b n ⁄ b , and n mod b b denotes the remainder of n n divided by b b .

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

f(10,87654)=8+7+6+5+4=30 f ( 10 ,   87654 ) = 8 + 7 + 6 + 5 + 4 = 30
f(100,87654)=8+76+54=138 f ( 100 ,   87654 ) = 8 + 76 + 54 = 138
You are given integers n n and s s . Determine if there exists an integer b(b≥2) b ( b ≥ 2 ) such that f(b,n)=s f ( b , n ) = s . If the answer is positive, also find the smallest such b b .

Constraints

1≤n≤1011 1 ≤ n ≤ 10 11
1≤s≤1011 1 ≤ s ≤ 10 11
n n , s s are integers.

Input

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

n
s

Output

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

---

题意

  给你一个n和一个s，问是否存在一个 b (2≤b) b   ( 2 ≤ b ) ，使得n在 b b 进制下各位数字的和为s，如果存在，则输出这个 b b ，如果不存在，则输出-1。

---

思路

  假设 x0,x1,…,xm x 0 , x 1 , … , x m 为n在 b b 进制下的各位数字（ x0 x 0 为最低位），那么有：

x0+x1·b+x2·b2+…xm·bm=n x 0 + x 1 · b + x 2 · b 2 + … x m · b m = n

x0+x1+x2+⋯+xm=s x 0 + x 1 + x 2 + ⋯ + x m = s

  考虑某个进制 b b ，使得n在 b b 进制下的表示的最高次幂大于等于 2 2 次幂，则有： b2≤n=b≤n‾√ b 2 ≤ n = b ≤ n ，于是我们在 b b 进制的最高次为2次幂及以上的时候直接暴力枚举即可。

  对于最高次为一次幂的 b b ，等式为：

x0+x1·b=n  ① x 0 + x 1 · b = n     ①

x0+x1=s  ② x 0 + x 1 = s     ②

  将 ② ② 式移项并代入 ① ① 式得：

s−x1+x1·b=n s − x 1 + x 1 · b = n

x1·(b−1)=n−s x 1 · ( b − 1 ) = n − s

  于是我们可以从大到小枚举 x1 x 1 （相当于从小到大枚举 (b−1) ( b − 1 ) ），判断一下是否存在合法 b b 的即可。

  因为前面我们枚举 b b 的时候已经枚举过 [2,n‾√] [ 2 , n ] 的范围，而 n−s‾‾‾‾‾√ n − s 显然小于 n‾√ n ，所以我们第二次枚举只用枚举 (n‾√,n−s] ( n , n − s ] 这个区间就可以了。

---

实现

#include 
long long n, s;
bool check(long long base) {
    long long ret = 0, tmp = n;
    while (tmp) ret += tmp % base, tmp /= base;
    return ret == s;
}
bool check(long long a1, long long tmp, long long a0 = 0, long long b = 0) {
    return (a0 = s - a1) >= 0 && a0 < (b = tmp / a1 + 1) && a1 < b;
}
int main() {
    scanf("%lld%lld", &n, &s);
    long long up = (long long)sqrt(n) + 1;
    for (long long base = 2; base <= up; base++) if (check(base)) return 0 * printf("%lld\n", base);
    if (n <= s) return 0 * printf("%lld\n", n < s ? -1: n + 1);
    long long tmp = n - s, upper = tmp / up + 1;
    for (long long a1 = upper; a1 >= 1; a1--)
        if (!(tmp % a1) && check(a1, tmp)) return 0 * printf("%lld\n", tmp / a1 + 1);
    return 0 * puts("-1");
}