【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+,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;

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

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

     return ;
 }