[英]Is there any way to reduce execution time of this code?
I am trying to find the sum of all the divisors of a given number But I am exceeding the time limit,help me to reduce the time limit of this code. 我正在尝试查找给定数字的所有除数的总和,但我超出了时间限制,请帮助我减少此代码的时间限制。
int a,count=0;
cin>>a;
for(int i=2;i<=a/2;i++) {
if(a%i==0) {
count=count+i;
}
}
count++;
cout<<count;
如果您一次将两个除数相加,则可以使循环运行到sqrt(a)
,而不是a / 2
: count += i + a / i
I would say go up to sqrt(a). 我会说上sqrt(a)。 Each time you have a remainder 0, add both the i and a/i.
每次余数为0时,都将i和a / i相加。 You will need to take care of the corner cases, but this should bring down the time complexity.
您将需要处理一些特殊情况,但这会降低时间的复杂性。 Depending on how large a is this should be faster.
根据a的大小,它应该更快。 For small values this may actually be slower.
对于较小的值,实际上可能会更慢。
This problem can be optimized by prime factorization . 这个问题可以通过素数分解来优化。
Let’s assume any number’s prime factor is = a ^n*b^m*c^k
Then, Total number of devisors will be = (n+1)(m+1)(k+1)
And sum of divisors = (a^(n+1) -1 )/(a-1) * (b^(m+1)-1)/(b-1) *(c^(k+1)-1)/(c-1)
X = 10 = 2^1 * 5^1
Total number of devisors = (1+1)(1+1) =2*2= 4
Sum of divisors = (2^2 – 1 ) /1 * (5^2 -1 )/4 = 3 * 24/4 = 18
1+2+5+10 = 18
Thanks everyone for the help..I got the answer 谢谢大家的帮助。
bool is_perfect_square(int n)
{
if (n < 0)
return false;
int root(round(sqrt(n)));
return n == root * root;
}
main()
{
int t;
cin>>t;
while(t--)
{
int a,count=0;
cin>>a;
bool c=is_perfect_square(a);
for(int i=2;i<=sqrt(a);i++)
{
if(a%i==0)
{
count=count+i+a/i;
}
}
if(c==true)
{
count = count - sqrt(a);
}
count++;
cout<<count<<endl;
}
}
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