【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 ;

 }