时间：2021-07-01 10:21:17 帮助过：22人阅读
B. Little Dima and Equation
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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?=?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
output
310 2008 13726
input
1 2 -18
output
input
2 2 -1
output
41 31 337 967
代码如下：
#include#include #include #include #include #include using namespace std;typedef __int64 LL;LL ss[10017];LL F(LL num){ LL s = 0; while(num) { s += num%10; num/=10; } return s;}LL Fac(LL num, LL n){ LL s = 1; for(LL i = 1; i <= n; i++) { s*=num; } return s;}int main(){ LL a, b, c; while(~scanf("%I64d%I64d%I64d",&a,&b,&c)) { memset(ss,0,sizeof(ss)); LL l = 0, i; for(i = 1; i <= 81; i++)//枚举 { LL x = b*Fac(i,a)+c; if(x >= 1000000000 || x <= 0) continue; LL ans = (int)F(x); if(ans == i) ss[l++] = x; } printf("%I64d\n",l); for(i = 0; i < l; i++) { if(i == 0) printf("%I64d",ss[i]); else printf(" %I64d",ss[i]); } printf("\n"); } return 0;}