HDU 5139数据离线处理

此题可以找到规律f(n) = 1! * 2! *...*n!, 如果直接打表的话，由于n比较大(10000000)，所以会超内存，这时候就要用到离线处理数据，就是先把数据存起来，到最后在暴力一遍求解就行了，代码如下

代码一(超内存):

 1 #include <stdio.h>
 2 
 3 const long long mod = 1000000007;
 4 const int N = 10000007;
 5 long long a[N];
 6 
 7 int main()
 8 {
 9     a[0] = a[1] = 1;
10     for (int i = 2; i < N; i++)
11     {
12         a[i] = a[i - 1] * i % mod;
13     }    
14     
15     for (int i = 2; i < N; i++)
16     {
17         a[i] = a[i - 1] * a[i] % mod;
18     }
19     int n;
20     while (~scanf("%d", &n))
21     {
22         printf("%I64d
", a[n]);
23     }
24     return 0;
25 }

代码二(AC)

 1 #include <stdio.h>
 2 #include <vector>
 3 #include <map>
 4 const long long mod = 1000000007;
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     int n;
10     vector<int> a;
11     map<int, long long> vis;
12     while (scanf("%d", &n) != EOF)
13     {
14         a.push_back(n);
15         vis[n] = -1;
16     }
17     long long prep = 1, nowp, prea = 1, nowa;
18     for (int i = 1; i <= 10000000; i++)//离线算法 
19     {
20         nowp = prep * i % mod;//相当于求阶乘 
21         prep = nowp;
22         nowa = nowp * prea % mod;//求n个阶乘的乘积 
23         prea = nowa;
24         if (vis.count(i))
25         {
26             vis[i] = nowa;
27         }
28     }    
29     for (int i = 0; i < a.size(); i++)//暴力一遍 
30     {
31         
32         printf("%I64d
", vis[a[i]]);
33     }
34     return 0;
35 }

离线算法确实挺神奇的，要深刻理解才能做题