hdu1215 七夕节---因子和

题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=1215
题目大意：
求N的因子和（不包括N本身）
解题思路：
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<queue>
 7 #include<stack>
 8 #include<map>
 9 #include<sstream>
10 #define Mem(a, b) memset(a, b, sizeof(a))
11 using namespace std;
12 typedef long long ll;
13 const int INF = 1e9 + 7;
14 const int maxn = 1000000+10;
15 int prime[maxn];
16 bool is_prime[maxn];
17 int sieve(int n)//返回n以内素数的个数
18 {
19     int p = 0;
20     for(int i = 0; i <= n; i++)is_prime[i] = 1;
21     is_prime[0] = is_prime[1] = 0;
22     for(ll i = 2; i <= n; i++)
23     {
24         if(is_prime[i])
25         {
26             prime[p++] = i;
27             for(ll j = i * i; j <= n; j += i)is_prime[j] = 0;//这里涉及i*i,必须使用long long
28         }
29     }
30     return p;
31 }
32 
33 ll Divisors_num(ll n, int tot)//素数总数
34 {
35     ll ans = 1;
36     for(int i = 0; i < tot && prime[i] * prime[i] <= n; i++)
37     {
38         if(n % prime[i] == 0)
39         {
40             int cnt = 0;
41             while(n % prime[i] == 0)
42             {
43                 cnt++;
44                 n /= prime[i];
45             }
46             ans *= (cnt + 1);
47         }
48     }
49     if(n > 1)ans *= 2;
50     return ans;
51 }
52 ll pow(ll a, ll b)
53 {
54     ll ans = 1;
55     while(b)
56     {
57         if(b & 1)ans = ans * a;
58         a *= a;
59         b /= 2;
60     }
61     return ans;
62 }
63 ll Divisors_sum(ll n, int tot)
64 {
65     ll ans = 1;
66     for(int i = 0; i < tot && prime[i] * prime[i] <= n; i++)
67     {
68         if(n % prime[i] == 0)
69         {
70             int cnt = 0;
71             while(n % prime[i] == 0)
72             {
73                 cnt++;
74                 n /= prime[i];
75             }
76             ans = (pow(prime[i], cnt + 1) - 1) / (prime[i] - 1) * ans;
77         }
78     }
79     if(n > 1)ans *= (n + 1);
80     return ans;
81 }
82 int main()
83 {
84     int T, cases = 0;
85     int tot = sieve(1000000);
86     cin >> T;
87     while(T--)
88     {
89         ll n;
90         scanf("%lld", &n);
91         printf("%lld
", Divisors_sum(n, tot) - n);
92     }
93     return 0;
94 }