【HDOJ6651】Final Exam（贪心）

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：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
#define N  1100000
#define M  4100000
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=1e9+7,inv2=(MOD+1)/2;
      double eps=1e-6;
      int INF=1e9;


ll read()
{
   ll v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}



int main()
{
    //freopen("1.in","r",stdin);
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        ll n=read(),m=read()+1,k=read();
        ll t=n-k+1;
        if(m%t) printf("%I64d
",m+(k-1)*(m/t+1));
         else printf("%I64d
",m+(k-1)*(m/t));
    }

    return 0;
}