「CF460B」Little Dima and Equation

2014年8月21日2,2340

Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following problem as a punishment.

Find all integer solutions x (0 < x < 109) of the equation:

x = b·s(x)a + c, where a, b, c are some predetermined constant values and function s(x) determines the sum of all digits in the decimal representation of number x.

The teacher gives this problem to Dima for each lesson. He changes only the parameters of the equation: a, b, c. Dima got sick of getting bad marks and he asks you to help him solve this challenging problem.

Input

The first line contains three space-separated integers: a, b, c (1 ≤ a ≤ 5; 1 ≤ b ≤ 10000; - 10000 ≤ c ≤ 10000).

Output

Print integer n — the number of the solutions that you’ve found. Next print n integers in the increasing order — the solutions of the given equation. Print only integer solutions that are larger than zero and strictly less than 109.

Sample test(s)

input

3 2 8

3 2 8

output

3
10 2008 13726

1 2	3 10 2008 13726

input

1 2 -18

1 2 -18

output

input

2 2 -1

2 2 -1

output

4
1 31 337 967

1 2	4 1 31 337 967

题解

直接枚举S（x）即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<cstdlib>
#define inf 1000000000
#define linf 9000000000000000000LL
#define ll long long
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int a,b,c,ans[100005];
int cal(int x)
{
    int sum=0;
    while(x)
    {
        sum+=x%10;
        x/=10;
    }
    return sum;
}
void solve()
{
    for(int i=1;i<=100;i++)
    {
        ll t=1;
        for(int j=1;j<=a;j++)
            t*=i;
        if(t*b+c>=1000000000)return;
        if(cal(t*b+c)==i)ans[++ans[0]]=t*b+c;
    }
}
int main()
{
    a=read();b=read();c=read();
    solve();
    printf("%d\n",ans[0]);
    for(int i=1;i<=ans[0];i++)
        printf("%d ",ans[i]);
    return 0;
}

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cmath>

#include<queue>

#include<vector>

#include<map>

#include<cstdlib>

#define inf 1000000000

#define linf 9000000000000000000LL

#define ll long long

using namespace std;

inline int read()

{

int x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

int a,b,c,ans[100005];

int cal(int x)

{

int sum=0;

while(x)

{

sum+=x%10;

x/=10;

}

return sum;

}

void solve()

{

for(int i=1;i<=100;i++)

{

ll t=1;

for(int j=1;j<=a;j++)

t*=i;

if(t*b+c>=1000000000)return;

if(cal(t*b+c)==i)ans[++ans[0]]=t*b+c;

}

int main()

{

a=read();b=read();c=read();

solve();

printf("%d\n",ans[0]);

for(int i=1;i<=ans[0];i++)

printf("%d ",ans[i]);

return 0;

}

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「CF460B」Little Dima and Equation

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