NYOJ 179 LK's problem ......

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LK's problem

时间限制：3000 ms  |  内存限制：65535 KB

难度：1

描述

LK has a question.Coule you help her?

It is the beginning of the day at a bank, and a crowd  of clients is already waiting for the entrance door to  open. 

Once the bank opens, no more clients arrive, and  tellerCount tellers begin serving the clients. A  

teller takes serviceTime minutes to serve each client.  clientArrivals specifies how long each client has  already been waiting at the moment when the bank door  opens. Your program should determine the best way to arrange the clients into tellerCount queues, so that  the waiting time of the client who waits longest is minimized. The waiting time of a client is the sum of  the time the client waited outside before the bank opened, the time the client waited in a queue once the  bank opened until the service began, and the service time of the client. Return the minimum waiting time for the client who waits the longest.

输入

The input will consist of several test cases. For each test case, one integer N (1<= N <= 100) is given in the first line. Second line contains N integers telling us the time each client had waited.Third line contains tow integers , teller's count and service time per client need. The input is terminated by a single line with N = 0.

输出

For each test of the input, print the answer.

样例输入

21 21 10110 50 500

样例输出

2160

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iphxer

/***类型：入门级题目**题目来源：NYOJ 179**时间：2017/7/22**解决方案：*/#include<stdio.h>#include<algorithm>using namespace std;int main(){int n, i, j, a, b;int c[110], d[110];while(scanf("%d",&n)&&n!=0){for(i = 0; i < n; i++){scanf("%d",&c[i]);}scanf("%d%d",&a,&b);sort(c,c+n);printf("%d ",c[0]);for(j=0,i=n-1;i>=0;i=i-a,j++){d[j] = c[i] + b*(j+1);}sort(d,d+j);printf("%d\n",d[j-1]); }  return 0;}//sort 从小到大排序，此题未理解题意