题意
给定 \(n,m,k\) (\(1\le n,m\le10^9\)，)问在 \(k\) 进制下有多少个分数值不同的 \(\frac{x}{y}\) 满足 \(x\le n,y\le m\) 且其小数形式的循环节从小数点后第一位开始。
sol
因为要求不同分数值，我们只考虑既约分数。类比十进制，故要求 \(gcd(y,k)=1\)。所以答案为：
\[\sum_{i=1}^{n}\sum_{j=1}^{m}[(i,j)=1][(j,k)=1] \]很明显需要进一步化简：
\[\begin{aligned} &\sum_{i=1}^{n}\sum_{j=1}^{m}[(i,j)=1][(j,k)=1]\\ =&\sum_{i=1}^{n}\sum_{j=1}^{m}[(i,j)=1]\sum_{d|(j,k)}\mu(d)\\ =&\sum_{d|k}\mu(d)\sum_{i=1}^{n}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}[(i,j)=1][(i,d)=1]\\ \end{aligned} \]可以发现有一部分和原本的式子长得比较像，设原答案为 \(f(n,m,k)\)，有
\[f(n,m,k)=\sum_{d|k}\mu(d)f(\lfloor\frac{m}{d}\rfloor,n, d) \]可以递归求解，边界为 \(f(0,m,k)=f(n,0,k)=0\)。
\[\begin{aligned} f(n,m,1)=&\sum_{i=1}^{n}\sum_{j=1}^{m}[(i,j)=1]\\ =&\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|(i,j)}\mu(d)\\ =&\sum_{d=1}^{min(n,m)}\mu(d)\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor \end{aligned} \]很经典的整除分块。但是这里 \(\mu\) 的前缀和要用到 \(10^9\) 所以使用杜教筛。
AC Code
#include <bits/stdc++.h>
namespace _default{
using namespace std;
#define lovely return
#define _lzy_ 0
#define LL long long
#define DB double
#define PII std::pair<int, int>
#define X first
#define Y second
#define uF(i, x, y) for(int i = (x); i <= (y); ++ i)
#define uf(i, x, y) for(int i = (x); i < (y); ++ i)
#define dF(i, x, y) for(int i = (x); i >= (y); -- i)
#define df(i, x, y) for(int i = (x); i > (y); -- i)
#define ef(i, u) for(int i = head[(u)]; i; i = ne[i])
#define sett(x, y) memset(x, y, sizeof x)
#define copy(x, y) memcpy(x, y, sizeof x)
const int MOD = 1e9 + 7, INF = 1e18;
const DB EPS = 1e-8;
template<typename T> inline T read(){
T s = 0; int f = 0; char ch = getchar();
while(!isdigit(ch)){if(ch == '-') f = 1; ch = getchar();}
while(isdigit(ch)) s = s * 10 + ch - 48, ch = getchar();
return f ? ~s + 1 : s;
}
template<typename T> inline void write(T x, const char *s = ""){
static int st[40]; int top = 0;
if(x < 0){putchar('-'); x = -x;}
if(!x) putchar('0');
while(x) st[++ top] = x % 10, x /= 10;
while(top) putchar(st[top --] + 48);
printf("%s", s);
}
template<typename T> inline void updmin(T &x, T y){if(x > y) x = y;}
template<typename T> inline void updmax(T &x, T y){if(x < y) x = y;}
template<typename T> inline void updadd(T &x, T y){(x += y % MOD) %= MOD;}
template<typename T> inline void updmul(T &x, T y){(x *= y % MOD) %= MOD;}
int cmp(DB a, DB b){if(fabs(a - b) < EPS) return 0; return a - b > EPS ? 1 : -1;}
}
using namespace _default;
const int N = 2005, M = 2e7;
int n, m, k;
int mu[M + 10], kk[M + 10], tot;
LL sm[M + 10];
bool Not_Prime[M + 10];
vector<int> p;
unordered_map<int, int> ttm;
struct lzy{
int n, m, k;
friend bool operator < (const lzy &a, const lzy &b){
if(a.n != b.n) return a.n < b.n;
else if(a.m != b.m) return a.m < b.m;
else return a.k < b.k;
}
};
map<lzy, LL> ans;
void xxs(){
mu[1] = 1;
uF(i, 2, M){
if(!Not_Prime[i]){
p.push_back(i);
mu[i] = -1;
}
for(auto j : p){
if(i * j > M) break;
Not_Prime[i * j] = true;
if(i % j == 0) break;
mu[i * j] = -mu[i];
}
}
uF(i, 1, M) sm[i] = sm[i - 1] + mu[i];
}
int getmu(int n){
if(n <= M) return sm[n];
if(ttm.count(n)) return ttm[n];
LL res = 1;
for(int l = 2, r; l <= n; l = r + 1){
r = n / (n / l);
res -= 1ll * (r - l + 1) * getmu(n / l);
}
return ttm[n] = res;
}
int Calc(int n, int m, int k){
if(!n || !m) return 0;
lzy tmp = (lzy){n, m, k};
if(ans.count(tmp)) return ans[tmp];
int res = 0;
if(k == 1){
int maxn = min(n, m);
for(int l = 1, r; l <= maxn; l = r + 1){
r = min(maxn, min(n / (n / l), m / (m / l)));
res += (getmu(r) - getmu(l - 1)) * (n / l) * (m / l);
}
ans[(lzy){m, n, k}] = res;
}
else for(int i = 1; i <= tot && kk[i] <= k; ++ i) if(k % kk[i] == 0)
res += Calc(m / kk[i], n, kk[i]) * mu[kk[i]];
return ans[tmp] = res;
}
signed main(){
n = read<int>(), m = read<int>(), k = read<int>();
xxs();
uF(i, 1, k) if(k % i == 0) kk[++ tot] = i;
write(Calc(n, m, k));
lovely _lzy_;
}