标签：codeforces acm algorithm 算法
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:
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.
The first line contains three space-separated integers: a,?b,?c (1?≤?a?≤?5; 1?≤?b?≤?10000; ?-?10000?≤?c?≤?10000).
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.
3 2 8
3 10 2008 13726
1 2 -18
0
2 2 -1
4 1 31 337 967
）。然后就对着式子敲出来就行了，注意long long和x的范围小于1e9，这两个地方是这道题的cha点，各种血腥的cha啊。我提交的过的代码没有判断小于1e9，然后我就天真的把lock了。呵呵呵呵呵，然后被cha了，呵呵呵呵呵呵。还好考5
2 100 cha到一个人挽回点损失~#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;
int d[100000];
int main()
{
int a,b,c,sum,cou;
long long temp,flag;
while(cin>>a>>b>>c)
{
memset(d,0,sizeof(d));
cou=0;
for(int i=1;i<=81;i++)
{
sum=0;
temp=1;
for(int j=0;j<a;j++)
{
temp*=i;
}
temp=b*temp;
temp=c+temp;
//printf("%d ",temp);
flag=temp;
while(temp)
{
sum+=temp%10;
temp/=10;
}
if(sum==i&&flag>0&&flag<(1e9))
d[cou++]=flag;
}
if(cou==0)
printf("0\n");
else
{
printf("%d\n%d",cou,d[0]);
for(int i=1;i<cou;i++)
{
printf(" %d",d[i]);
}
printf("\n");
}
}
}
Codeforces Round #262 (Div. 2) B,布布扣,bubuko.com
Codeforces Round #262 (Div. 2) B
标签：codeforces acm algorithm 算法
原文地址：http://blog.csdn.net/notdeep__acm/article/details/38724963
