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 = 1000000007;
const int N = 10000007;
long long a[N];

int main()
{
    a[0] = a[1] = 1;
    for (int i = 2; i < N; i++)
    {
        a[i] = a[i - 1] * i % mod;
    }    
    
    for (int i = 2; i < N; i++)
    {
        a[i] = a[i - 1] * a[i] % mod;
    }
    int n;
    while (~scanf("%d", &n))
    {
        printf("%I64d
", a[n]);
    }
    return 0;
} 

代码二(AC)

#include <stdio.h>
#include <vector>
#include <map>
const long long mod = 1000000007;
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] = -1;
    }
    long long prep = 1, nowp, prea = 1, nowa;
    for (int i = 1; i <= 10000000; 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 = 0; i < a.size(); i++)//暴力一遍 
    {
        
        printf("%I64d
", vis[a[i]]);
    }
    return 0;
}

离线算法确实挺神奇的，要深刻理解才能做题