高精度-HDU-1042-N!

N!
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 68606 Accepted Submission(s): 19629

Problem Description
Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!

Input
One N in one line, process to the end of file.

Output
For each N, output N! in one line.

Sample Input

1
2
3

Sample Output

1
2
6

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题意：还需要解释么。。。1W以内的阶乘运算。

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题解：一开始想着是不是打个表，后面MLE了几次，哭死，最后也懒得本地打表，索性把打表撤了一次一次算，耗时有点长，不过还是A了。

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#include <iostream>#include <string.h>#include <stdlib.h>#include <stdio.h>#include <algorithm>#include <math.h>using namespace std;struct BigInt{    const static int mod = 10000;    const static int DLEN = 4;    int a[10000],len;    BigInt()    {        memset(a,0,sizeof(a));        len = 1;    }    BigInt(int v)    {        memset(a,0,sizeof(a));        len = 0;        do        {            a[len++] = v%mod;            v /= mod;        }        while(v);    }    BigInt(const char s[])    {        memset(a,0,sizeof(a));        int L = strlen(s);        len = L/DLEN;        if(L%DLEN)len++;        int index = 0;        for(int i = L-1; i >= 0; i -= DLEN)        {            int t = 0;            int k = i - DLEN + 1;            if(k < 0)k = 0;            for(int j = k; j <= i; j++)                t = t*10 + s[j] - '0';            a[index++] = t;        }    }    BigInt operator +(const BigInt &b)const    {        BigInt res;        res.len = max(len,b.len);        for(int i = 0; i <= res.len; i++)            res.a[i] = 0;        for(int i = 0; i < res.len; i++)        {            res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);            res.a[i+1] += res.a[i]/mod;            res.a[i] %= mod;        }        if(res.a[res.len] > 0)res.len++;        return res;    }    BigInt operator *(const BigInt &b)const    {        BigInt res;        for(int i = 0; i < len; i++)        {            int up = 0;            for(int j = 0; j < b.len; j++)            {                int temp = a[i]*b.a[j] + res.a[i+j] + up;                res.a[i+j] = temp%mod;                up = temp/mod;            }            if(up != 0)                res.a[i + b.len] = up;        }        res.len = len + b.len;        while(res.a[res.len - 1] == 0 &&res.len > 1)res.len--;        return res;    }    void output()    {        printf("%d",a[len-1]);        for(int i = len-2; i >=0 ; i--)            printf("%04d",a[i]);        printf("\n");    }};BigInt a,temp;int main(int argc, const char * argv[]){    a=BigInt(1);    int n;    while(cin >> n)    {        a=BigInt(1);        for(int i=2;i<=n;i++)        {            temp=BigInt(i);            a=a*temp;        }        a.output();    }    return 0;}