題目描述
輸入一個正整數N,輸出N的階乘。
輸入描述:
正整數N(0<=N<=1000)
輸出描述:
輸入可能包括多組數據,對於每一組輸入數據,輸出N的階乘
示例1
輸入
4 5 15
輸出
24 120 1307674368000
//C++實現
#include <cstdio>
#include <cstring>
using namespace std;
struct bign{
int d[100000];
int len;
bign(){
memset(d,0,sizeof(d));
len=0;
}
};
bign change(char str[]){
bign a;
a.len=strlen(str);
for(int i=0;i<a.len;i++){
a.d[i]=str[a.len-i-1]-'0';
}
return a;
}
bign muti(bign a,int b){
bign c;
int carry=0;
for(int i=0;i<a.len;i++){
int temp=a.d[i]*b+carry;
c.d[c.len++]=temp%10;
carry=temp/10;
}
while(carry!=0){
c.d[c.len++]=carry%10;
carry/=10;
}
return c;
}
bign fun(int n){
bign sum;
sum.d[0]=1;
sum.len=1;
while(n!=1){
sum=muti(sum,n);
n--;
}
return sum;
}
void print(bign a){
for(int i=a.len-1;i>=0;i--){
printf("%d",a.d[i]);
}
printf("\n");
}
int main(){
int n;
while(scanf("%d",&n)!=EOF){
bign sum;
sum=fun(n);
print(sum);
}
return 0;
}
//java實現
import java.math.BigInteger;
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int number=sc.nextInt();
System.out.println(fun(number));
}
}
public static BigInteger fun(int number){
if(number==1){
return BigInteger.valueOf(1);
}else
return fun(number-1).multiply(BigInteger.valueOf(number));
}
}