623 - 500!
Time limit: 3.000 seconds
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=564
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
完整代码：
/*0.172s*/
#include<cstdio>
int factorial[1001][2800];
int main(void)
{
int i, j, n;
factorial[0][0] = 1;
factorial[1][0] = 1;
for (i = 2; i < 1001; i++)
for (j = 0; j < 2800; j++)
{
factorial[i][j] += factorial[i - 1][j] * i;
if (factorial[i][j] > 9)
{
factorial[i][j + 1] = factorial[i][j] / 10;
factorial[i][j] %= 10;
}
}
while (~scanf("%d", &n))
{
printf("%d!\n", n);
for (i = 2600; i > 0; i--)
if (factorial[n][i])
break;
for (; i >= 0; i--)
printf("%d", factorial[n][i]);
putchar('\n');
}
return 0;
}
查看本栏目更多精彩内容：http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/
01./*1.238s*/ 02. 03.import java.io.*; 04.import java.util.*; 05.import java.math.*; 06. 07.public class Main { 08. static final int maxn = 1001; 09. static Scanner cin = new Scanner(new BufferedInputStream(System.in)); 10. 11. public static void main(String[] args) { 12. BigInteger[] f = new BigInteger[maxn]; 13. f[0] = BigInteger.ONE; 14. for (int i = 1; i < maxn; ++i) 15. f[i] = f[i - 1].multiply(BigInteger.valueOf(i)); 16. while (cin.hasNextInt()) { 17. int n = cin.nextInt(); 18. System.out.println(n + "!"); 19. System.out.println(f[n]); 20. } 21. } 22.}
以上是小编为您精心准备的的内容，在的博客、问答、公众号、人物、课程等栏目也有的相关内容，欢迎继续使用右上角搜索按钮进行搜索close() scanner java
， for
， number
， biginteger
， of
， should
， 100的阶乘
， 1135
which
高精度阶乘、高精度阶乘算法、高精度阶乘和、高精度算法 求n的阶乘、高精度求阶乘，以便于您获取更多的相关知识。