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):

 #include <stdio.h>

 const long long mod = ;
 const int N = ;
 long long a[N];

 int main()
 {
     a[] = a[] = ;
     for (int i = ; i < N; i++)
     {
         a[i] = a[i - ] * i % mod;
     }    

     for (int i = ; i < N; i++)
     {
         a[i] = a[i - ] * a[i] % mod;
     }
     int n;
     while (~scanf("%d", &n))
     {
         printf("%I64d\n", a[n]);
     }
     return ;
 } 

代码二(AC)

 #include <stdio.h>
 #include <vector>
 #include <map>
 const long long mod = ;
 using namespace std;

 int main()
 {
     int n;
     vector<int> a;
     map<int, long long> vis;
     while (scanf("%d", &n) != EOF)
     {
         a.push_back(n);
         vis[n] = -;
     }
     long long prep = , nowp, prea = , nowa;
     for (int i = ; i <= ; i++)//离线算法
     {
         nowp = prep * i % mod;//相当于求阶乘
         prep = nowp;
         nowa = nowp * prea % mod;//求n个阶乘的乘积
         prea = nowa;
         if (vis.count(i))
         {
             vis[i] = nowa;
         }
     }
     for (int i = ; i < a.size(); i++)//暴力一遍
     {

         printf("%I64d\n", vis[a[i]]);
     }
     return ;
 }

离线算法确实挺神奇的，要深刻理解才能做题