HDU 1018

题意：求n！的位数

题解：斯特林数: log10(n!)=1.0/2*log10(2*pi*n)+n*log10(n/e)。

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 #include <math.h>
 5 #define E 2.71828182
 6 #define PI acos( -1.0 )
 7 
 8 int n;
 9 
10 int main()
11 {
12     int T;
13     scanf( "%d", &T );
14     while( T-- )
15     {
16         scanf( "%d", &n );
17         double sum = (double)n*log10( n / E ) + 0.5*log10( 2.0*PI*n );
18         printf( "%d
", (int)sum+1 );
19     }
20     return 0;
21 }

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当然这题也可以用暴力。

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 #include <math.h>
 5 
 6 int n;
 7 
 8 int main()
 9 {
10     int T;
11     scanf( "%d", &T );
12     while( T-- )
13     {
14         scanf( "%d", &n );
15         double sum = 0;
16         for( int i = 1; i <= n; ++i )
17             sum += log10( (double)i );
18         printf( "%d
", (int)sum+1 );
19     }
20     return 0;
21 }

View Code