codeforces Gym 100500C D.Hall of Fame 排序-白红宇

codeforces Gym 100500C D.Hall of Fame 排序

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发布时间：2019-06-26

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Hall of Fame

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100500/attachments

Description

It was the second day of IBM Chill Zone, and it was the time for distributing the prizes. Unfortunately due to unknown reasons, the organizing committee can only buy T-shirts to the contestants or give them D dollars in cash. The T-shirts factory only permitted them to order a single bulk of T-shirts of the same size where a single T-shirt price was determined using this equation: C × S Where C is a given constant, and S is the size of the T-shirt. There was only one limitation on the T-shirts they will give to the contestants, any contestant only accepts to receive a T-shirt of greater or equal size but not smaller size. The organizing committee decided to minimize the money they should pay . Please help them in determining the amount they need to pay in order to give each contestant either a T-shirt or D dollars.

Input

T the number of test cases. For each test case there will be three integers N, D and C , then N integers representing the size of each of the T-shirt of each of the contestants. 1 ≤ T ≤ 100 1 ≤ N ≤ 100, 000 1 ≤ D ≤ 10, 000 1 ≤ C ≤ 10, 000 1 ≤ t − shirtsize ≤ 100, 000

Output

For each test case print a single line containing: Case_x:_y x is the case number starting from 1. y is is the required answer. Replace the underscores with spaces.

Sample Input

1 5 100 1 35 70 75 90 110

Sample Output

Case 1: 425

HINT

题意

你可以给人买衣服，也可以给他D元，衣服的价格等于衣服的大小乘以布料钱

求最少给多少钱

题解：

排序，然后O(n)扫一遍就好啦~

代码

#include 
     
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                   typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define test freopen("test.txt","r",stdin)const int maxn=202501;#define mod 1000000007#define eps 1e-9const int inf=0x3f3f3f3f;const ll infll = 0x3f3f3f3f3f3f3f3fLL;inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}//*************************************************************************************ll a[maxn];int main(){ int t=read(); for(int cas=1;cas<=t;cas++) { ll n=read(),d=read(),c=read(); for(int i=1;i<=n;i++) a[i]=read(); sort(a+1,a+n+1); ll ans=infll; for(int i=1;i<=n;i++) { ans=min(ans,a[i]*c*(i)+(n-i)*d); } ans=min(ans,n*d); printf("Case %d: %lld\n",cas,ans); }}

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