C algorithm problem, please point out with high fingers,

#include""It's the same as yours!

The above is roughly correct, the multiplication symbol is wrong, hehe.

C++ Algorithm Questions:

According to the meaning of the title, n<=2*10 9

In the red box, **i*j in the figure may exceed this range, resulting in the maximum overflow of integers not getting the correct result. It should be changed to a for loop, starting with the largest number and searching for the smaller number, not multiplication, but division.

for(int i = n-1; n >=sqrt(n); i--)if(n%i==0)

maxnum= i;

break;

Of course, these must ensure that the input n must be satisfied, and only the multiplication of two prime numbers equals n, otherwise it is necessary to increase the judgment whether it is a prime number.

Your ** complexity is too high.

Roughly estimated, o(sqrt(n)*n).

n data range [0,10 9], definitely timed out.

In general programming problems, the algorithm with a data range of more than 10 6, o(n) may time out.

I wrote a slightly faster one, with a complexity of about o(sqrt(n)*sqrt(i)), which should be less than o(n) you try.

#include

using namespace std;

bool isp(int x)

for(int i=2;i*i<=x;i++)return true;

int main()

return 0;

It's not that there is no output, it's that the efficiency of this algorithm is relatively low, the amount of calculation is too large, as long as you are willing to wait, the result will still be produced, but the time of this waiting will be larger with the number, the longer the time will be.

I've changed my mind, and the time will be significantly shortened. Let's experience it for yourself below.

=The following is ** ==

One of the test files failed like this: input 1999520027, output 99991 but after I input the 1999520027, there is no output result? Don't know why?

#include

#include

#include

#include

using namespace std;

vector array_a,array_b;

int isprime(int n)

if(bit_sum % 3)

coutreturn 0;

Obviously, the macro define base3 (base2+(rate*(break3-break2))) is correct. define base3 (base1+base2+(rate3*(break3-break2))) is a macro that actually adds base1 twice, which doesn't make sense from any point of view.

The third picture is correct. In the second diagram, the 360kWh fee for the first level is double-counted (because this fee is already included in base2).