公式题。。。
本福特定律说明在b进位制中，以数n起头的数出现的机率为logb(n + 1) − logb(n) .
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 1205
#define maxm 40005
#define eps 1e-10
#define mod 10000007
#define INF 1e9
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
//head
int main(void)
{
int _, __, n, q, b;
while(scanf("%d", &_)!=EOF) {
__ = 0;
while(_--) {
scanf("%d%d%d", &n, &b, &q);
double x = (log(n+1) - log(n))/log(10.0);
if(q == 1) {
while(b > n) b /= 10;
if(b == n) x = 1.0;
else x = 0.0;
printf("Case #%d: %.5f\n", ++__, x);
}
else if(q == 10 || q == 100 || q == 1000) {
b *= 10000;
while(b > n) b /= 10;
if(b == n) x = 1.0;
else x = 0.0;
printf("Case #%d: %.5f\n", ++__, x);
}
else printf("Case #%d: %.5f\n", ++__, x);
}
}
return 0;
}
