Problem:
Score : 1000 points

Problem Statement

In a two-dimensional plane, we have a rectangle R whose vertices are (0,0), (W,0), (0,H), and (W,H), where W and H are positive integers.
Here, find the number of triangles \Delta in the plane that satisfy all of the following conditions:

Each vertex of \Delta is a grid point, that is, has integer x- and y-coordinates.

\Delta and R shares no vertex.

Each vertex of \Delta lies on the perimeter of R, and all the vertices belong to different sides of R.

\Delta contains at most K grid points strictly within itself (excluding its perimeter and vertices).

Constraints

1 \leq W \leq 10^5

1 \leq H \leq 10^5

0 \leq K \leq 10^5

Input

Input is given from Standard Input in the following format:

W H K

Output

Print the answer.

Sample Input 1

2 3 1

Sample Output 1

12

For example, the triangle with the vertices (1,0), (0,2), and (2,2) contains just one grid point within itself and thus satisfies the condition.

Sample Input 2

5 4 5

Sample Output 2

132

Sample Input 3

100 100 1000

Sample Output 3

461316

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <unistd.h>
#include <sys/time.h>
#include <stdlib.h>
 
using namespace std;
 
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<P, ll> T;
typedef pair<long double , ll> Ps;
 
const ll INF = 2e18;
const ll fact_table = 3200008;
long double Pi = 3.1415926535897932384626;

priority_queue <ll> pql;
priority_queue <P> pqp;
priority_queue <P> bag;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop
 
ll dx[8]={1,0,-1,0,1,1,-1,-1};
ll dy[8]={0,1,0,-1,1,-1,-1,1};
char dir[] = "DRUL";
//↓,→,↑,←

#define endl "\n"
#ifdef ENJAPMA
    #undef endl
#endif
#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<x<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define CLEAR(a) a = decltype(a)(); 
#define pc(x) cout << x << ",";
#define rep(i, n) for (ll i = 0; i < (n); i ++)

// const ll mod = 998244353ll;
const ll mod = 1000000007ll;

ll mypow(ll a, ll b, ll mod){ll x=1;while(b){while(!(b&1)){(a*=a)%=mod;b>>=1;}(x*=a)%=mod;b--;}return x;}
void YES(bool cond){ if(cond){ p("YES");}else{ p("NO");} return;}
void Yes(bool cond){ if(cond){ p("Yes");}else{ p("No");} return;}

/*
ll fact[fact_table + 5],rfact[fact_table + 5]; 
 
void c3_init(){
    fact[0] = rfact[0] = 1;
    for(ll i=1; i<=fact_table; i++){
        fact[i] = (fact[i-1]*i) % mod;
    }
    rfact[fact_table] = mypow(fact[fact_table],mod - 2, mod);
    for(ll i=fact_table; i>=1; i--){
       rfact[i-1] = rfact[i] * i;
       rfact[i-1] %= mod;
    }
    return;}
ll c3(ll n,ll r){
    return (((fact[n] * rfact[r]) % mod ) * rfact[n-r]) % mod;}
*/

void init(){
    struct timeval _time;
    gettimeofday(&_time, NULL);
    ll usec = _time.tv_usec * 1000000;
    srand(usec);
    return;
}

ll range(ll a,ll b){
    // generate random number [a, b]
    // if b >= 10^8, this might not be precise
    ll num = rand() % (1 + b - a) + a;
    return num;
}

bool multicase = false;
ll mygcd(ll a, ll b){
    if(b > a)swap(a, b);
    if(b == 0 || a == b)return a;
    return mygcd(b, a % b);
}

ll calc(ll w, ll h, ll k){
    ll ans = 0;
    for(ll i1=1;i1<h;i1++){
        for(ll i2=1;i2<h;i2++){
            if(min(i1, i2) * w > 4 * k)break;
            for(ll j=1;j<w;j++){
                ll b = mygcd(i1, j) + mygcd(w - j, i2) + mygcd(w, abs(i1 - i2));
                ll s = max(i1, i2) * w * 2;
                s -= (i1 * j) + (w - j) * (i2) + (abs(i1 - i2) * w);
                if(s - b + 2 <= 2 * k)ans ++;
            }
        }
    }
    return ans;
}
ll n,m,num,a,b,c,d,e,h,q;
void solve(){
    ll w, k;
    cin >> w >> h >> k;
    ll ans = 0;
    ans += calc(w, h, k) * 2;
    ans += calc(h, w, k) * 2;
    p(ans);
    return;
}

int main(){
    // init();
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll q, testcase = 1;
    if(multicase){
        cin >> q;
    }else{
        q = 1; 
    }
    while(q--){
        // pu("Case ");pu("#");pu(testcase);pu(": ");
        solve();
        testcase++;
    }
    // solve();
    return 0;
}