1 /*
 2 1：log10(12345)=log10(1.2345*10^4)=4+log(1.2345);n的位数就是log10(n)+1;可以暴力。
 3 2：斯特林公式：一个数的阶乘近似等于sqrt(2*PI*n)*(n/e)^n;
 4 */
 5 #include<iostream>
 6 #include<cmath>
 7 #include<cstdio>
 8 using namespace std;
 9 int main()
10 {
11     int n,m;
12     scanf("%d",&n);
13     while(n--)
14     {
15         scanf("%d",&m);
16         double ans=0;
17         for(int i=2;i<=m;i++)
18         ans+=log10(i);
19         printf("%d\n",(int)ans+1);
20     }
21     return 0;
22 }
23 
24 #include<iostream>
25 #include<cmath>
26 #include<cstdio>
27 using namespace std;
28 const double e=2.718281828459;
29 const double PI=3.14159265;
30 int main()
31 {
32     int n,m;
33     scanf("%d",&n);
34     while(n--)
35     {
36         double ans;
37         scanf("%d",&m);
38         if(m!=1)
39         ans=0.5*log10(2*PI*m)+m*log10(m)-m*log10(e)+1;
40         else ans=1.0;
41         printf("%d\n",(int)ans);
42     }
43     return 0;
44 }

//两个大数相加用字符串处理。
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
int main()
{
    int t;
    char s1[1005],s2[1005];
    scanf("%d",&t);
    for(int k=1;k<=t;k++)
    {
        scanf("%s%s",s1,s2);
        int ks1=strlen(s1);
        int ks2=strlen(s2);
        ks1--;ks2--;
        int sav=0,h=0,a1,a2;
        char s[1005];
        while(1)
        {
            if(ks1<0&&ks2<0)
            break;
            if(ks1>=0&&ks2>=0)
            {
                a1=s1[ks1]-‘0‘;
                a2=s2[ks2]-‘0‘;
            }
            if(ks1>=0&&ks2<0)
            {
                a1=s1[ks1]-‘0‘;
                a2=0;
            }
            if(ks1<0&&ks2>=0)
            {
                a1=0;
                a2=s2[ks2]-‘0‘;
            }
            ks1--;ks2--;
            int tem=a1+a2+sav;
            sav=tem/10;
            tem%=10;
            s[h++]=tem+‘0‘;
        }
        if(sav!=0)
        s[h++]=sav+‘0‘;
        printf("Case %d:\n",k);
        printf("%s + %s = ",s1,s2);
        for(int i=h-1;i>=0;i--)
        cout<<s[i];
        printf("\n");
        if(k!=t)
        printf("\n");
    }
    return 0;
}