HDU 1299 Diophantus of Alexandria

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/*1/x+1/y=1/n

give you an integer n to find the number of x, y combinations (n<=10^9)

x, y two variables in the formula can be converted into one condition

Suppose y=n+m; the expression of x is x=n*n/m+n;

The condition is if (n*n%m==0)
So the answer is the number of n*n factors.
Obviously, the general method will be timed out directly.
But the factor of n is seldom the maximum is sqrt (n);
Each number can represent the product of the element factor
N= (Prime[0]^ans[0]) * ((prime[1]^ans[1]) * ....
Knowing the number of each factor, the other non-element factors of n can be combined by the element factor.
So sum= (1+ans[0]) * (1+ans[1]) ...//, this kind of combination of the mother function,1 means no, we just need his coefficient

Because N*n= ((prime[0]^ (ans[0]*2)) * ((prime[1]^ (ans[1]*2)) * ....
sum= (1+ans[0]*2) * (1+ans[1]*2) ... * *

must transform the formula into a condition to find an unknown expression.
Suppose y=n+m; the expression of x is x=n*n/m+n;
So the answer is the number of n*n factors.
Obviously, it's going to time out.
But the n factor is rarely sqrt (n);
Each number can represent the product of the element factor
N= (Prime[0]^ans[0]) * ((prime[1]^ans[1]) * ....
I know the number of each element factor.
1 means no so sum= (1+ans[0]) * (1+ans[1]) ...
Because N*n= ((prime[0]^ (ans[0]*2)) * ((prime[1]^ (ans[1]*2)) * ....

sum= (1+ans[0]*2) * (1+ans[1]*2) ... * *

#include <stdio.h> #include <string.h> #define LL __int64#define N 330000//The largest element factor is sqrt (n) bool Used[n];int Prime[n];int cnt;void init () {    memset (used,false,sizeof (used));    for (int i=2;i<n;i++)    {        if (!used[i])        {            prime[cnt++]=i;//filter element factor for        (int j=i;j<n;j+=i)        {            used[j]=true;}}}    } int main () {    int t,k=1;    cnt=0;    Init ();    scanf ("%d", &t);    while (t--)    {        ll n,sum=1;        scanf ("%i64d", &n);        for (int i=0;i<cnt;i++)        {            ll ans=0;            if (prime[i]>n) break                ;            while (n%prime[i]==0)            {                ans++;                N/=prime[i];            }            Sum= (2*ans+1) *sum;        }        if (n>1)//n itself is a prime word            sum*=3;        printf ("Scenario #%d:\n", k++);        printf ("%i64d\n\n", sum/2+1);    }    return 0;}

HDU 1299 Diophantus of Alexandria