HDU - 1042 N!(高精度BigInt)

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 86335    Accepted Submission(s): 25398

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

 

Author

JGShining（極光炫影）

 

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#include <bits/stdc++.h>
using namespace std;

const int N = 1e4 + 10;
int n;
struct BigInt
{
    const static int Mod = 10000;
    const static int LEN = 4;
    int a[N], len;
    BigInt(){
        memset(a, 0, sizeof a);
        len = 1;
    }
    void init(int x){
        memset(a, 0, sizeof a);
        len = 0;
        do{
            a[len++] = x%Mod;
            x /= Mod;
        }while(x);
    }
    int Compare(const BigInt &b){  /// a>b 1, a=b 0, a<b -1
        if(len < b.len) return -1;
        if(len > b.len) return 1;
        for(int i = len-1; i >= 0; i--)
            if(a[i] < b.a[i])return -1;
            else if(a[i] > b.a[i])return 1;
        return 0;
    }
    BigInt operator +(const BigInt &b)const{
        BigInt ans;
        ans.len = max(len, b.len);
        for(int i = 0; i <= ans.len; i++) ans.a[i] = 0;
        for(int i = 0; i< ans.len; i++){
            ans.a[i] += ((i < len) ? a[i] : 0)+((i < b.len) ? b.a[i] : 0);
            ans.a[i+1] += ans.a[i]/Mod;
            ans.a[i] %= Mod;
        }
        if(ans.a[ans.len] > 0)ans.len++;
        return ans;
    }
    BigInt operator -(const BigInt &b)const{
        BigInt ans;
        ans.len = len;
        int k = 0;
        for(int i = 0; i < ans.len; i++){
            ans.a[i] = a[i]+k-b.a[i];
            if(ans.a[i] < 0) ans.a[i] += Mod,k = -1;
            else k = 0;
        }
        while(ans.a[ans.len-1] == 0 && ans.len > 1) ans.len--;
        return ans;
    }
    BigInt operator *(const BigInt &b)const{
        BigInt ans;
        for(int i = 0; i < len;i++){
            int k = 0;
            for(int j = 0; j < b.len; j++){
                int temp = a[i]*b.a[j]+ans.a[i+j]+k;
                ans.a[i+j] = temp%Mod;
                k = temp/Mod;
            }
            if(k != 0) ans.a[i+b.len] = k;
        }
        ans.len = len+b.len;
        while(ans.a[ans.len-1] == 0 && ans.len > 1)ans.len--;
        return ans;
    }
    void output(){
        printf("%d", a[len-1]);
        for(int i = len-2; i >= 0; i--)
            printf("%04d", a[i]);
        printf("\n");
    }
};

int main(){
    while(scanf("%d", &n) == 1){
        BigInt ans, x;
        ans.init(1);
        for(int i = 2;i <= n; i++){
            x.init(i);
            ans = ans*x;
        }
        ans.output();
    }
}