HDU 1999 Non-touch number (number N of the Sieve method to find the sum of the true factors) __ACM

Topic address
The main idea: S (n) is the sum of the true factors of n of positive integers, a factor that is less than n and divisible by n, such as S (12) =1+2+3+4+6=16. If any number m,s (m) is not equal to N, then it is said that N is a non-touch number, and that the number within 1000 is not a number.
To solve the problem: to determine whether a number n is not a number, we must determine whether there is not a positive integer m of the sum of the true factor, it will be within 1000 of the sum of the true factor to find the sum of the true factor and the Sieve method

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <string>


using namespace std;


const int MAXN = 1000000+10;
const int MAXM = 1000+10;


int SUM[MAXN],SIGN[MAXM];


int main ()
{
    memset (sum,0,sizeof (sum));
    memset (sign,0,sizeof (sign));
    for (int i = 1; I <= 500000; i++)/sieve method to find S (n) and save it with sum of the array
    {for
        (int j = 2*i; J <= 1000000; j = i)
        {
            SUM[J] + = i
        }
    }
    for (int i = 1; I <= 1000000 i++)//1000 of S (m)
        if (Sum[i] < 1000)
            sign[sum[i] = 1;
    int t,n;
    scanf ("%d", &t);
    while (t--)
    {
        scanf ("%d", &n);
        if (!sign[n])
            puts ("yes");
        Else
            puts ("no");
    return 0;
}