N! HDU 1042

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60244    Accepted Submission(s): 17166

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

 

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
using namespace std;
int a[8000];
int main()
{
    int n;
    int i,j;
    while(~scanf("%d",&n))
    {
        if(n==0)
        {
            printf("1
");
            continue;
        }
        memset(a,0,sizeof(a));
        a[0]=1;
        for(i=2;i<=n;i++)
        {
            for(j=0;j<8000;j++)
            a[j]*=i;
            for(j=0;j<8000;j++)
            {
                a[j+1]+=a[j]/100000;
                a[j]%=100000;
            }
        }
        for(i=8000;i>=0;i--)
        {
            if(a[i]!=0)
            break;
        }
        printf("%d",a[i--]);
        for(;i>=0;i--)
        printf("%05d",a[i]);
        printf("
");
    }
    return 0;
}
-----------------------------------------------
#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
using namespace std;
int a[100000];
int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    int i,j,temp;
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        a[0]=1;
        int count=1;
        for(i=1;i<=n;i++)
        {
            int k=0;
            for(j=0;j<count;j++)
            {
                temp = a[j]*i+k;
                a[j]=temp%10;
                k=temp/10;
            }
            while(k)
            {
                a[count++]=k%10;
                k/=10;
            }
        }
        for(i=100000;i>=0;i--)
        {
            if(a[i]!=0)
            break;
        }
        for(;i>=0;i--)
        printf("%d",a[i]);
        printf("
");
    }
    return 0;
}

 模板：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <assert.h>
#include <ctype.h>
#include <map>
#include <string>
#include <set>
#include <bitset>
#include <utility>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <iostream>
#include <fstream>
#include <list>
using  namespace  std;

const  int MAXL = 100000;
struct  BigNum
{
    int  num[MAXL];
    int  len;
    BigNum()
    {
        memset(num,0,sizeof(num));
    }
};

//高精度比较 a > b return 1, a == b return 0; a < b return -1;
int  Comp(BigNum &a, BigNum &b)
{
    int  i;
    if(a.len != b.len) return (a.len > b.len) ? 1 : -1;
    for(i = a.len-1; i >= 0; i--)
        if(a.num[i] != b.num[i]) return  (a.num[i] > b.num[i]) ? 1 : -1;
    return  0;
}

//高精度加法
BigNum  Add(BigNum &a, BigNum &b)
{
    BigNum c;
    int  i, len;
    len = (a.len > b.len) ? a.len : b.len;
    memset(c.num, 0, sizeof(c.num));
    for(i = 0; i < len; i++)
    {
        c.num[i] += (a.num[i]+b.num[i]);
        if(c.num[i] >= 10)
        {
            c.num[i+1]++;
            c.num[i] -= 10;
        }
    }
    if(c.num[len])
        len++;
    c.len = len;
    return  c;
}
//高精度乘以低精度，当b很大时可能会发生溢出int范围,具体情况具体分析
//如果b很大可以考虑把b看成高精度
BigNum Mul1(BigNum &a, int  &b)
{
    BigNum c;
    int  i, len;
    len = a.len;
    memset(c.num, 0, sizeof(c.num));
    //乘以0，直接返回0
    if(b == 0)
    {
        c.len = 1;
        return  c;
    }
    for(i = 0; i < len; i++)
    {
        c.num[i] += (a.num[i]*b);
        if(c.num[i] >= 10)
        {
            c.num[i+1] = c.num[i]/10;
            c.num[i] %= 10;
        }
    }
    while(c.num[len] > 0)
    {
        c.num[len+1] = c.num[len]/10;
        c.num[len++] %= 10;
    }
    c.len = len;
    return  c;
}

//高精度乘以高精度,注意要及时进位，否则肯能会引起溢出,但这样会增加算法的复杂度，
//如果确定不会发生溢出, 可以将里面的while改成if
BigNum  Mul2(BigNum &a, BigNum &b)
{
    int i, j, len = 0;
    BigNum  c;
    memset(c.num, 0, sizeof(c.num));
    for(i = 0; i < a.len; i++)
    {
        for(j = 0; j < b.len; j++)
        {
            c.num[i+j] += (a.num[i]*b.num[j]);
            if(c.num[i+j] >= 10)
            {
                c.num[i+j+1] += c.num[i+j]/10;
                c.num[i+j] %= 10;
            }
        }
    }
    len = a.len+b.len-1;
    while(c.num[len-1] == 0 && len > 1)
        len--;
    if(c.num[len])
        len++;
    c.len = len;
    return  c;
}
void  print(BigNum &a)   //输出大数
{
    int  i;
    for(i = a.len-1; i >= 0; i--)
        printf("%d", a.num[i]);
    puts("");
}

void Init(BigNum &a, char *s, int &tag)   //将字符串转化为大数
{
    int  i = 0, j = strlen(s);
    if(s[0] == '-')
    {
        j--;
        i++;
        tag *= -1;
    }
    a.len = j;
    for(; s[i] != ''; i++, j--)
        a.num[j-1] = s[i]-'0';
}

int main(void)
{
    int n;
    while(~scanf("%d",&n))
    {
        BigNum a;
        int tag=1;
        Init(a,"1",tag);
        for(int i=1;i<=n;i++)
        {
            a=Mul1(a,i);
        }
        print(a);
    }
    return 0;
}