题意：有n门课，价值之和为m，每门课的价值可能是0到m
一门价值为x的课需要花至少x+1时间准备才能通过
问不管价值如何分配都能通过至少k门课的最小总准备时间
m，n，k<=1e9
思路：
1 #include<bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 typedef unsigned int uint;
5 typedef unsigned long long ull;
6 typedef pair<int,int> PII;
7 typedef pair<ll,ll> Pll;
8 typedef vector<int> VI;
9 typedef vector<PII> VII;
10 #define N 1100000
11 #define M 4100000
12 #define fi first
13 #define se second
14 #define MP make_pair
15 #define pi acos(-1)
16 #define mem(a,b) memset(a,b,sizeof(a))
17 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
18 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
19 #define lowbit(x) x&(-x)
20 #define Rand (rand()*(1<<16)+rand())
21 #define id(x) ((x)<=B?(x):m-n/(x)+1)
22 #define ls p<<1
23 #define rs p<<1|1
24
25 const ll MOD=1e9+7,inv2=(MOD+1)/2;
26 double eps=1e-6;
27 int INF=1e9;
28
29
30 ll read()
31 {
32 ll v=0,f=1;
33 char c=getchar();
34 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
35 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
36 return v*f;
37 }
38
39
40
41 int main()
42 {
43 //freopen("1.in","r",stdin);
44 int cas;
45 scanf("%d",&cas);
46 while(cas--)
47 {
48 ll n=read(),m=read()+1,k=read();
49 ll t=n-k+1;
50 if(m%t) printf("%I64d\n",m+(k-1)*(m/t+1));
51 else printf("%I64d\n",m+(k-1)*(m/t));
52 }
53
54 return 0;
55 }