1 #include <stdio.h>
 2 #include <math.h>
 3 #include <queue>
 4 #include <vector>
 5 #include <stack>
 6 #include <map>
 7 #include <string>
 8 #include <cstring>
 9 #include <algorithm>
10 #include <iostream>
11 #define maxe 500000
12 int a[500010];
13 int main()
14 {
15     int T,i,j;
16     scanf("%d",&T);
17     memset(a,0,sizeof(a));
18     for(i=1;i<=maxe;i++)
19     {
20         for(j=2;i*j<=maxe;j++)
21         {
22            a[i*j]=a[i*j]+i;
23         }
24     }
25     int n;
26     while(T--)
27     {
28        scanf("%d",&n);
29        printf("%d\n",a[n]);
30     }
31     return 0;
32
33 }

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<string>
 5 #include<cmath>
 6 #include<algorithm>
 7 using namespace std;
 8
 9 #define CLR(arr, what) memset(arr, what, sizeof(arr))
10 const int N = 500002;
11
12 int prim[N];
13 bool visit[N];
14
15 void fast_prim() //素数打表
16 {
17     memset(visit, true, sizeof(visit));
18     int limit = (int)sqrt(N * 1.0);
19     for(int i = 2; i < limit; ++i)
20         if(visit[i])
21             for(int j = i * i; j < N; j += i)
22                 visit[j] = false;
23     for(int i = 2, j = 0; i < 10000; ++i)
24         if(visit[i])
25             prim[j++] = i;
26 }
27
28 int prim_reduce(int n) //整数素分解
29 {
30     int limit, num, sum, total = 1, temp = n;
31     limit = sqrt(N * 1.0);
32     for(int i = 0; prim[i] * prim[i] <= n; ++i)
33     {
34         num = sum = 1;
35         if(n == 1)
36             break;
37         while(n % prim[i] == 0)
38         {
39             num *= prim[i];
40             n /= prim[i];
41             sum += num;
42         }
43             total *= sum;
44     }
45     if(n != 1)
46         total *= (n + 1);
47     return total - temp;
48 }
49
50 int main()
51 {
52     fast_prim();
53     int ncase;
54     int num;
55     scanf("%d", &ncase);
56     while(ncase--)
57     {
58         scanf("%d", &num);
59         if(num == 1)
60         {
61             printf("0\n");
62             continue;
63         }
64         else if(visit[num])
65         {
66             printf("1\n");
67             continue;
68         }
69         else
70             printf("%d\n", prim_reduce(num));
71     }
72     return 0;
73 }