题目:http://codeforces.com/contest/1418/problem/A
设需要a次1操作,则等式为 1+a*(x-1)=tot=k+k*y, a次1操作得到足够的s,再进行k次2操作,所以答案是a+k
化简为:
这里需要用到向上取整公式
#include<bits/stdc++.h> using namespace std; typedef long long ll; typedef double db; #define rep(i,a,b) for(ll i=a;i<=b;i++) #define endl ' ' #define V vector #define pb(x) push_back(x); const ll amn=1e5+5; ll a[amn]; int main(){ // cout<<ceil(0.1); ios::sync_with_stdio(0); ll T=1; cin>>T; while(T--){ ll x,y,k; cin>>x>>y>>k; cout<<(k*(y+1)-1+x-1-1)/(x-1)+k<<endl; } }