gcj Round 1- A -B题

输入

E R N

v1 v2 ... vn

有n个工作，没做一个消耗一定的能量，但可以得到R个能量（最多为E，多则去除），初始能量是E， 每件工作的收益是

v[i]*u[i], u[i] 做i使用的能量. u[i]>=0, u[i]<=E

解：

如果规定u[i]>=1,那么这个问题就很简单， find the max in V[ ], then use E energy to do it, the others use R to do them.

So max gain is Vmax*E+  sum(Vnon-max *R).

But u[i] can be zero.

that is for a position i, the energy before can be accumulated, if u[i] is much bigger then then all the j before i, but u[i] is not the maximum of V[].

Like

E =5 R=1

V[]=  1 1 1 99 1 1 100

1*1 + 1*1 +1*1 +1*99 + 1*1 +1*1 +5*100 < 1*1 + 1*1 + 1*1 +99*3 + 0*1 +0*1 +5*100

 

u[0]=E

for(i=1;i

 u[i]=R;

for(j=i-1;j>=0;j--){

   if(v[j]>=v[i]) break;

 if(u[j]+u[i]<=E){ u[i]=u[i]+u[j]; u[j]=0; }

else {

u[j]=u[j]-(E-u[i]);

u[i]=E;

break;
}

}

sum=0;

for(i=0;i

 

转载于:https://www.cnblogs.com/gaoqichao/archive/2013/05/01/3053765.html