题目要求 (frac{x}{y}) 是纯循环，如果记 ({x}) 为 (x) 的小数部分，那么就存在 (l) 满足：
[{ frac{x}{y} } = { frac{x k^l}{y} }
]
这就要求 (x equiv xk^l (mod ; y))。又 (gcd(x, y) = 1)，我们得到 (gcd(k, y) = 1)
我们的答案就是
[egin{split} Ans &= sum_{x = 1} ^ n sum_{y = 1} ^ m [gcd(x, y) = 1][gcd(y, k) = 1] \ &= sum_{y = 1} ^ m [gcd(y, k) = 1]sum_{x = 1} ^ n sum_{d|x\d|y} mu(d) \ &= sum_{d = 1} ^ n mu(d) lfloor frac{n}{d}
floor sum_{d|y} ^ m [gcd(y,k) = 1] \ &= sum_{d = 1} ^ n mu(d) lfloor frac{n}{d}
floor sum_{i = 1} ^ {lfloor frac{m}{d}
floor}[gcd(id,k) = 1] \ &= sum_{d = 1} ^ n mu(d) lfloor frac{n}{d}
floor [gcd(d,k) = 1] sum_{i = 1} ^ {lfloor frac{m}{d}
floor}[gcd(i,k) = 1] end{split}
]
令 (f(n) = sum_{i = 1} ^ n [gcd(i, k) = 1])，则 $ f(n) = lfloor frac{n}{k} floor varphi(k) + f(n mod k)$，可以预处理后 (O(1)) 计算。
[Ans = sum_{d = 1} ^ n lfloor frac{n}{d}
floor f(lfloor frac{m}{d}
floor) mu(d) [gcd(d,k) = 1]
]
对这个式子整除分块，现在就只需要计算 (g(n) = mu(n)[gcd(i,k) = n]) 的前缀和，记为 (h(n))。
[egin{split} h(n) &= sum_{i = 1} ^ {n} g(i) \ &= sum_{i = 1} ^ n mu(i) - sum_{d = 2, d | k} ^ n sum_{i = 1} ^ n [gcd(i,k) = d] mu(i) \ &= sum_{i = 1} ^ n mu(i) - sum_{d = 2, d | k} ^ n sum_{i = 1} ^ {lfloor frac{n}{d}
floor} [gcd(i,frac{k}{d}) = 1] mu(id) \ &= sum_{i = 1} ^ n mu(i) - sum_{d = 2, d | k} ^ n sum_{i = 1} ^ {lfloor frac{n}{d}
floor} mu(i) mu(d) [gcd(i,frac{k}{d}) = 1] [gcd(i,d) = 1] \ &= sum_{i = 1} ^ n mu(i) - sum_{d = 2, d | k} ^ n mu(d) sum_{i = 1} ^ {lfloor frac{n}{d}
floor} mu(i) [gcd(i,k) = 1] \ &= sum_{i = 1} ^ n mu(i) - sum_{d = 2, d | k} ^ n mu(d) h(lfloor frac{n}{d}
floor) end{split}
]
$ sum_{i = 1} ^ n mu(i)$ 可以用杜教筛算。
复杂度懒得分析了，随便调了个块大小过了。
#pragma GCC optimize("2,Ofast,inline")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define LL long long
#define pii pair<int, int>
using namespace std;
const int Maxn = 1e6;
const int N = 1e6 + 10;
const int inf = 0x3f3f3f3f;
template <typename T> T read(T &x) {
int f = 0;
register char c = getchar();
while (c > '9' || c < '0') f |= (c == '-'), c = getchar();
for (x = 0; c >= '0' && c <= '9'; c = getchar())
x = (x << 3) + (x << 1) + (c ^ 48);
if (f) x = -x;
return x;
}
LL n, m, k, tot;
int miu[N], ispri[N], pri[N], phi[N];
int f[N], smiu[N];
vector<int> divi;
struct Hashtable {
static const int MOD = 1e7 + 7;
int num[MOD], val[MOD];
void ins(int x, int v) {
int t = x % MOD;
while (num[t] && num[t] != x) t = (t + 1 == MOD) ? 0 : t + 1;
num[t] = x;
val[t] = v;
}
int find(int x) {
int t = x % MOD;
while (num[t] && num[t] != x) t = (t == 0) ? MOD - 1 : t - 1;
return val[t];
}
} Map, h;
int gcd(int x, int y) {
return (!y) ? x : gcd(y, x % y);
}
void prework() {
miu[1] = 1;
phi[1] = 1;
for (int i = 2; i <= Maxn; ++i) {
ispri[i] = 1;
}
for (int i = 2; i <= Maxn; ++i) {
if (ispri[i]) {
pri[++tot] = i;
miu[i] = -1;
phi[i] = i - 1;
}
for (int j = 1; j <= tot && i * pri[j] <= Maxn; ++j) {
ispri[i * pri[j]] = 0;
if (i % pri[j] != 0) {
miu[i * pri[j]] = -miu[i];
phi[i * pri[j]] = phi[i] * (pri[j] - 1);
}
else {
miu[i * pri[j]] = 0;
phi[i * pri[j]] = phi[i] * pri[j];
break;
}
}
}
for (int i = 1; i <= k; ++i) {
f[i] = f[i - 1] + (gcd(i, k) == 1);
}
for (int i = 1; i <= Maxn; ++i) {
smiu[i] = smiu[i - 1] + miu[i];
}
for (int i = 1; i <= k; ++i) {
if (k % i == 0) divi.pb(i);
}
memset(h.val, 0x3f, sizeof h.val);
}
inline int F(int n) {
return n / k * phi[k] + f[n % k];
}
inline int nex(int n, int i) {
return n / (n / i);
}
int calc_miu(int x) {
if (x <= Maxn) return smiu[x];
int ans;
if (ans = Map.find(x)) return ans;
ans = 1;
for (int i = 2; i <= x; ++i) {
int j = nex(x, i);
ans -= (j - i + 1) * calc_miu(x / i);
i = j;
}
Map.ins(x, ans);
return ans;
}
int calc_h(LL x) {
int S;
if ((S = h.find(x)) != inf) return S;
S = calc_miu(x);
for (int i = 1; i < divi.size(); ++i) {
if (divi[i] > x) break;
S -= miu[divi[i]] * calc_h(x / divi[i]);
}
h.ins(x, S);
return S;
}
int main() {
read(n); read(m); read(k);
prework();
LL ans = 0;
for (int i = 1; i <= n && i <= m; ++i) {
int j = min(nex(n, i), nex(m, i));
LL tmp = 1LL * (n / i) * F(m / i);
LL sumg = calc_h(j) - calc_h(i - 1);
ans += tmp * sumg;
i = j;
}
cout << ans << endl;
return 0;
}