题解

\(f(n)\)的取值范围最多\(9^2 * 18\)

直接枚举判断就好

代码

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define pdi pair<db,int>

#define mp make_pair

#define pb push_back

#define enter putchar('\n')

#define space putchar(' ')

#define eps 1e-8

#define mo 974711

#define MAXN 500005

//#define ivorysi

using namespace std;

typedef long long int64;

typedef double db;

template<class T>

void read(T &res) {

    res = 0;char c = getchar();T f = 1;

    while(c < '0' || c > '9') {

	if(c == '-') f = -1;

	c = getchar();

    }

    while(c >= '0' && c <= '9') {

	res = res * 10 + c - '0';

	c = getchar();

    }

    res *= f;

}

template<class T>

void out(T x) {

    if(x < 0) {x = -x;putchar('-');}

    if(x >= 10) {

	out(x / 10);

    }

    putchar('0' + x % 10);

}

int64 k,a,b;

int Calc(int64 t) {

    int res = 0;

    while(t) {

	res += (t % 10) * (t % 10);

	t /= 10;

    }

    return res;

}

void Solve() {

    read(k);read(a);read(b);

    int cnt = 0;

    for(int i = 1 ; i <= 2000 ; ++i) {

	if(b / i < k) break;

	int64 t = k * i;

	if(a <= t && t <= b) {

	    if(Calc(t) == i) cnt++;

	}

    }

    out(cnt);enter;

}

int main() {

#ifdef ivorysi

    freopen("f1.in","r",stdin);

#endif

    Solve();

    return 0;

}