answer
I feel this is the hardest problem first conclusion.
x / y co-prime first, then if the decimal is pure decimals, is not difficult to think of y is not a multiple of 2.5 just fine.
Because Decimal is divided by 2 and 5 are in addition to have the best.
Bound extra something.
If k is binary, it can be compared to give gcd (y, k) = 1.
prove:
Loop is assumed that the number of bits of pure l
The \ (x * k ^ l \ equiv x (mod ~ y) \)
\(k^l\equiv 1(mod~y)\)
L to exist, then it must have \ (gcd (k, the y-) = 1 \) , turn the same.
Inversion:
\(Ans=\sum_{i=1}^n\sum_{j=1}^m[(i,j)=1]*[(j,k)=1]\))
\(\sum_{d=1}^{min(n,m)} \mu(d)*[gcd(d,k)=1]*(n/d)*\sum_{j=1}^{m/d}[gcd(j,k)=1]\)
If the prefix can solve the foregoing problems and, after receiving block behind repellent or pretreated O (1) is not a problem.
Then you found in front of a multiplicative function, and can be min_25 screen multiplicative function, but to a screen twice.
Gone ......
Code：
#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, B = y; i <= B; i ++)
#define ff(i, x, y) for(int i = x, B = y; i < B; i ++)
#define fd(i, x, y) for(int i = x, B = y; i >= B; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
using namespace std;
int k;
int d[20], d0;
void fen(int x) {
for(int i = 2; i * i <= x; i ++) if(x % i == 0) {
d[++ d0] = i;
while(x % i == 0) x /= i;
}
if(x > 1) d[++ d0] = x;
}
namespace m25 {
const int N = 2e5 + 5;
int n, sq;
int w[N], w0, i1[N], i2[N];
int p[N], p0, bz[N];
void sieve(int n) {
fo(i, 2, n) {
if(!bz[i]) p[++ p0] = i;
for(int j = 1; i * p[j] <= n; j ++) {
int k = i * p[j];
bz[k] = 1; if(i % p[j] == 0) break;
}
}
}
int g[N];
#define num(x) ((x) <= sq ? i1[x] : i2[n / (x)])
int sp[N];
void build() {
sq = sqrt(n); sieve(sq);
for(int i = 1, j; i <= n; i = j + 1) {
j = n / (n / i);
w[++ w0] = n / i;
if(w[w0] <= sq) i1[w[w0]] = w0; else i2[i] = w0;
g[w0] = w[w0] - 1;
}
fo(j, 1, p0) for(int i = 1; p[j] * p[j] <= w[i] && i <= w0; i ++) {
int k = num(w[i] / p[j]);
g[i] -= (g[k] - (j - 1));
}
int l = 0;
fd(i, w0, 1) {
while(l < d0 && d[l + 1] <= w[i]) l ++;
g[i] -= l;
}
fo(i, 1, w0) g[i] = -g[i];
fo(i, 1, p0) sp[i] = sp[i - 1] - (k % p[i] != 0);
for(int j = p0; j >= 1; j --) if(k % p[j]) for(int i = 1; p[j] * p[j] <= w[i] && i <= w0; i ++) {
int k = num(w[i] / p[j]);
g[i] += (g[k] - sp[j]) * -1;
}
fo(i, 1, w0) g[i] ++;
}
ll q1(int x) {
return g[num(x)];
}
}
using m25 :: q1;
namespace m26 {
const int N = 2e5 + 5;
int n, sq;
int w[N], w0, i1[N], i2[N];
int p[N], p0, bz[N];
void sieve(int n) {
fo(i, 2, n) {
if(!bz[i]) p[++ p0] = i;
for(int j = 1; i * p[j] <= n; j ++) {
int k = i * p[j];
bz[k] = 1; if(i % p[j] == 0) break;
}
}
}
int g[N];
#define num(x) ((x) <= sq ? i1[x] : i2[n / (x)])
int sp[N];
void build() {
sq = sqrt(n); sieve(sq);
for(int i = 1, j; i <= n; i = j + 1) {
j = n / (n / i);
w[++ w0] = n / i;
if(w[w0] <= sq) i1[w[w0]] = w0; else i2[i] = w0;
g[w0] = w[w0] - 1;
}
fo(j, 1, p0) for(int i = 1; p[j] * p[j] <= w[i] && i <= w0; i ++) {
int k = num(w[i] / p[j]);
g[i] -= (g[k] - (j - 1));
}
int l = 0;
fd(i, w0, 1) {
while(l < d0 && d[l + 1] <= w[i]) l ++;
g[i] -= l;
}
fo(i, 1, w0) g[i] = -g[i];
fo(i, 1, p0) sp[i] = sp[i - 1] - (k % p[i] != 0);
for(int j = p0; j >= 1; j --) if(k % p[j]) for(int i = 1; p[j] * p[j] <= w[i] && i <= w0; i ++) {
int k = num(w[i] / p[j]);
g[i] += (g[k] - sp[j]) * -1;
}
fo(i, 1, w0) g[i] ++;
}
ll q2(int x) {
return g[num(x)];
}
}
using m26 :: q2;
int n, m;
struct P {
int x, y;
} t[1024]; int t0;
void dg(int x, int y, int z) {
if(x > d0) {
t[++ t0] = (P) {y, z};
return;
}
dg(x + 1, y, z);
dg(x + 1, y * d[x], -z);
}
int cmp(P a, P b) { return a.x < b.x;}
void query() {
ll ans = 0;
int lc = 0, c = 0;
for(int i = 1, j; i <= min(n, m); i = j + 1) {
int m1 = n / (n / i), m2 = m / (m / i);
if(m1 < m2) {
j = m1, c = 0;
} else {
j = m2, c = 1;
}
ll s = !c ? q1(j) : q2(j);
if(i > 1) s -= !lc ? q1(i - 1) : q2(i - 1);
lc = c;
ll s2 = 0;
fo(u, 1, t0) if(t[u].x <= m / i) s2 += t[u].y * (m / i / t[u].x);
ans += s * (n / i) * s2;
}
pp("%lld", ans);
}
int main() {
scanf("%d %d %d", &n, &m, &k);
fen(k);
m25 :: n = n;
m25 :: build();
m26 :: n = m;
m26 :: build();
dg(1, 1, 1);
sort(t + 1, t + t0 + 1, cmp);
query();
}