C :

#include<stdio.h>
int main()
{
    long long x,y,k;
    scanf("%lld%lld%lld",&x,&y,&k);
    //k个火把需要k个木棒k个煤炭
    //交易2需要交易k次能有k个煤炭
    //分析交易1 要知道总木棒数量
    //交易2就需要y*k个木棒
    //那么总共需要y*k+k个木棒
    //原本一个木棒就还需要y*k+k-1个木棒
    //初始化交易次数是k
    long long jiaoyi=k;
    long long jiaoyi1=(y*k+k-1)/(x-1);
    if((y*k+k-1)%(x-1)!=0)
    {
        jiaoyi1++;
    }
    jiaoyi=jiaoyi+jiaoyi1;
    printf("%lld",jiaoyi);
    return 0;
}

C++ :

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e5+10;
typedef pair<int,int> PII;
typedef priority_queue<int , vector<int>, greater<int>> minqueue;  //从小到大 queue
typedef priority_queue<int, vector<int>, less<int>> maxqueue;      //从大到小 queue

int gcd(int a, int b){  //最大公因数 
    return b ? gcd(b, a % b) : a;
}

int lcm(int a, int b){ //最小公倍数 
    return a * b / __gcd(a, b);
}

int qmi(int base, int power, int p)  //快速幂求余 
{
	int result = 1;   
	while (power > 0)           
	{
		if (power & 1)         							
			result = result * base % p;   
		base = base * base % p ;       						
		power >>= 1;         						
	}
	return result % p;
}

void solve()
{
	int x,y,k;
	cin>>x>>y>>k;
	int res = k+y*k - 1;
	int ans = k;
	x--;
	if(res%x!=0)
	{
		ans++;
	}
	ans += res / x;
	cout<<ans;
}

signed main()
{
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int t;
	t=1;
	//cin>>t; 
	while(t--)     
	{
		solve();
	}
	return 0;
}