Problem 30 Digit fifth powers （暴力枚举）

Digit fifth powers

Problem 30

Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:

1634 = 1⁴ + 6⁴ + 3⁴ + 4⁴
8208 = 8⁴ + 2⁴ + 0⁴ + 8⁴
9474 = 9⁴ + 4⁴ + 7⁴ + 4⁴

As 1 = 1⁴ is not a sum it is not included.

The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

Answer:

443839

Completed on Sat, 29 Oct 2016, 03:12

题解：如果是一个6位数，各位数字5次方之和最大为 6 * 95 = 354294，是一个6位数，所以6位数是有可能满足的。
如果是一个7位数，各位数字5次方之和最大为 7 * 95 = 413343，是一个6位数，不可能等于7位数自身，所以7位数是不可能满足的。注意：As 1 = 1⁴ is not a sum it is not included..

代码：

#include<bits/stdc++.h>
int sum(int n)
{
int sum=0;
int digit = 0;
while(n!=0)
{
digit = n%10;
sum += (int)pow((float)digit,5);
n /= 10;
}
return sum;
}
int main()
{
int i;
int ans = 0;
for (i=10;i<999999;i++)
{
if (sum(i) == i)
{
ans += i;
}
}
std::cout<<ans<<std::endl;
return 0;
}