思路:

把给定的数分解质因子之后,对于每一个质因子x,都需要x次操作(一次copy all操作和x-1次paste),所以答案就是对分解后的所有质因子求和。

实现:

 class Solution
{
public:
int minSteps(int n)
{
int sum = ;
for (int i = ; i <= n; i++)
{
while (n % i == )
{
sum += i;
n /= i;
}
}
return sum;
}
};

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