In these days you can more and more often happen to see programs which perform some useful calculations being executed rather then trivial screen savers. Some of them check the system message queue and in case of finding
it empty (for examples somebody is editing a file and stays idle for some time) execute its own algorithm.
As an examples we can give programs which calculate primary numbers.
One can also imagine a program which calculates a factorial of given numbers. In this case it is the time complexity of order O(n) which makes troubles, but the memory requirements. Considering the fact that 500! gives 1135-digit number no
standard, neither integer nor floating, data type is applicable here.
Your task is to write a programs which calculates a factorial of a given number.
Assumptions: Value of a number ``n" which factorial should be calculated of does not exceed 1000 (although 500! is the name of the problem, 500! is a small limit).
Input
Any number of lines, each containing value ``n" for which you should provide value of n!
Output
2 lines for each input case. First should contain value ``n" followed by character `!'.
The second should contain calculated value n!.
Sample Input
10
30
50
100
Sample Output
10!
3628800
30!
265252859812191058636308480000000
50!
30414093201713378043612608166064768844377641568960512000000000000
100!
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
題目大意
求n!,n的範圍是1-1000,只需要輸出即可。
思路
1.用數組存這個大數,模擬乘法的進位。下標0是低位,比如f[6][2],f[6][1],f[6][0]分別是720,即6!=720。
2.或者用Java的BigInteger類,不用加包名,類名寫成Main
代碼參考:https://blog.csdn.net/synapse7/article/details/11859307
BigInteger的簡單使用:https://blog.csdn.net/lsk1998/article/details/83057662
C++:
#include <stdio.h>
using namespace std;
int f[1010][3010];
void init()
{
f[0][0] = 1;
f[1][0] = 1;
for(int i=2;i<=1001;i++){
for(int j=0;j<=3000;j++){
f[i][j] += f[i-1][j] * i;
f[i][j+1] += f[i][j]/10;
f[i][j] = f[i][j]%10;
}
}
}
int main(){
init();
int n;
while(~scanf("%d",&n))
{
printf("%d!\n",n);
int m=3000;
while(f[n][m]==0) m--;
for(int i=m;i>=0;i--){
printf("%d",f[n][i]);
}printf("\n");
}
return 0;
}
Java:
import java.math.BigInteger;
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
while (cin.hasNext())
{
BigInteger num=new BigInteger("1");
BigInteger flat=new BigInteger("1");
int n=cin.nextInt();
for (int i = 0; i < n; i++) {
flat=flat.multiply(num);
num=num.add(new BigInteger("1"));
}
System.out.println(n+"!\n"+flat);
}
}
}