nyoj 28 nyoj 69-大数阶乘

http://acm.nyist.net/JudgeOnline/problem.php?pid=28 大数水题~~

 1 #include<stdio.h>
 2 int a[20000];
 3 void ch(int *a, int n)
 4 {
 5     int i, k;
 6     for(i = 19999; i >= 0; i--)
 7     {
 8         if(a[i])
 9             break;
10     }
11     k = i;
12     for(i = 0; i <= k; i++)
13     {
14         a[i] = a[i] * n;
15     }
16     for(i = 0; i < 20000; i++)
17     {
18         if(a[i] > 9)
19         {
20             a[i+1] += a[i] / 10;
21             a[i] = a[i] % 10;
22         }
23     }
24 }
25 int main()
26 {
27     int n, i;
28     a[0] = 1;
29     scanf("%d", &n);
30     for(i = 1; i <= n; i++)
31         ch(a, i);
32     for(i = 19999; i >= 0; i--)
33     {
34         if(a[i])
35             break;
36     }
37     for(; i >= 0; i--)
38         printf("%d", a[i]);
39     putchar(10);
40     return 0;
41 }

 这是具体计算阶乘的~

如果只需要计算大数阶乘的位数，像http://acm.nyist.net/JudgeOnline/problem.php?pid=69，有公式可以套~~~ n!的位数 = log10(2*PI*n)/2+n*log10(n/e);

代码如下：

 1 #include<iostream>
 2 #include<math.h>
 3 using namespace std;
 4 int main()
 5 {
 6     int tcases, n;
 7     cin >> tcases;
 8     while(tcases--)
 9     {
10         cin >> n;
11         if(n == 1)
12             cout << 1 << endl;
13         else
14         {
15             int result = (int) (log10(4.0 * acos(0.0) * n)/2 + 1.0 * n * log10(1.0 *n / exp(1.0))) + 1;  //arccos（0）= PI/2
16             cout << result << endl;
17         }
18     }
19     return 0;
20 }