Time limit : 2sec / Memory limit : 256MB

Score : 100 points

Problem Statement

There is a hotel with the following accommodation fee:

  • X yen (the currency of Japan) per night, for the first K nights
  • Y yen per night, for the (K+1)-th and subsequent nights

Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.

Constraints

  • 1≤N,K≤10000
  • 1≤Y<X≤10000
  • N, K, X, Y are integers.

Input

The input is given from Standard Input in the following format:

N
K
X
Y

Output

Print Tak's total accommodation fee.


Sample Input 1

Copy
5
3
10000
9000

Sample Output 1

Copy
48000

The accommodation fee is as follows:

  • 10000 yen for the 1-st night
  • 10000 yen for the 2-nd night
  • 10000 yen for the 3-rd night
  • 9000 yen for the 4-th night
  • 9000 yen for the 5-th night

Thus, the total is 48000 yen.


Sample Input 2

Copy
2
3
10000
9000

Sample Output 2

Copy
20000

题解:水过
 #include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#include <queue>
#include <stack>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll gcd(ll a,ll b){
return b?gcd(b,a%b):a;
}
bool cmp(int x,int y)
{
return x>y;
}
const int N=;
const int mod=1e9+;
int main()
{
std::ios::sync_with_stdio(false);
int n,k,x,y;
cin>>n>>k>>x>>y;
ll s=;
if(n<=k) s+=n*x;
else s+=k*x+(n-k)*y;
cout<<s<<endl;
return ;
}

AtCoder Beginner Contest 044 A - 高橋君とホテルイージー / Tak and Hotels (ABC Edit)的更多相关文章

  1. AtCoder Beginner Contest 044 C - 高橋君とカード / Tak and Cards

    题目链接:http://abc044.contest.atcoder.jp/tasks/arc060_a Time limit : 2sec / Memory limit : 256MB Score ...

  2. 高橋君とホテル / Tak and Hotels

    高橋君とホテル / Tak and Hotels Time limit : 3sec / Stack limit : 256MB / Memory limit : 256MB Score : 700  ...

  3. AtCoder Beginner Contest 045 B - 3人でカードゲームイージー / Card Game for Three (ABC Edit)

    Time limit : 2sec / Memory limit : 256MB Score : 200 points Problem Statement Alice, Bob and Charlie ...

  4. 高橋君とカード / Tak and Cards

    高橋君とカード / Tak and Cards Time limit : 2sec / Stack limit : 256MB / Memory limit : 256MB Score : 300 p ...

  5. 高橋君とカード / Tak and Cards AtCoder - 2037 (DP)

    Problem Statement Tak has N cards. On the i-th (1≤i≤N) card is written an integer xi. He is selectin ...

  6. AtCoder Beginner Contest 044 B - 美しい文字列 / Beautiful Strings

    Time limit : 2sec / Memory limit : 256MB Score : 200 points Problem Statement Let w be a string cons ...

  7. AtCoder Beginner Contest 022 A.Best Body 水题

    Best Body Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://abc022.contest.atcoder.jp/tasks/abc02 ...

  8. AtCoder Beginner Contest 177 题解

    AtCoder Beginner Contest 177 题解 目录 AtCoder Beginner Contest 177 题解 A - Don't be late B - Substring C ...

  9. AtCoder Beginner Contest 173 题解

    AtCoder Beginner Contest 173 题解 目录 AtCoder Beginner Contest 173 题解 A - Payment B - Judge Status Summ ...

随机推荐

  1. pip安装提示pkg_resources.DistributionNotFound: pip==18.1

    在用pip install安装依赖的时候提示pkg_resources.DistributionNotFound: pip==18.1,更新一下pip就可以了 easy_install pip==18 ...

  2. vc让界面保持最上层

    vc让界面保持最上层.事实上就一个函数就ok了, ::SetWindowPos(AfxGetMainWnd()->m_hWnd,HWND_TOPMOST,-1,-1,-1,-1,0);

  3. CGPoint,CGSize,CGRect转NSString以及CGRect的一些便捷实用方法

    打印代码小技巧 UIKIT_EXTERN NSString *NSStringFromCGPoint(CGPoint point); UIKIT_EXTERN NSString *NSStringFr ...

  4. 浅谈JS的变量提升

    JS的解析机制,是JS的又一大重点知识点,在面试题中更经常出现,今天就来唠唠他们的原理.首先呢,我们在我们伟大的浏览器中,有个叫做JS解析器的东西,它专门用来读取JS,执行JS.一般情况是存在作用域就 ...

  5. vue【指令】

    <div class="m-conbox"> <div v-text="html"></div> <div>{{ ...

  6. mysql 分组 列转行

    SELECT aa.type,CONCAT('(',GROUP_CONCAT('\'',aa.user_id separator '\'\,'),'\')') FROM (select  aa.typ ...

  7. maven设置每次构建获取最新版本号

    build.gradle中的依赖是通过设置maven依赖实现.我们知道,maven可以说是通过一个坐标定位来确定唯一一个包的,所说的坐标定位分别是groupId,artifactId和version三 ...

  8. [Java in NetBeans] Lesson 05. Method/function

    这个课程的参考视频和图片来自youtube. 主要学到的知识点有: Define a method:(motivation: write one time, but use it many times ...

  9. nodejs+mysql入门实例(改)

    //连接数据库 var mysql = require('mysql'); var connection = mysql.createConnection({ host: 'bdm253137448. ...

  10. android studio 编译sdk版降低报错解决方法

    解决办法如下: 步骤1:在gradle中修改 compile sdk 版本,比如 8. 步骤2:在gradle中删除v7包的依赖 步骤3:在manifest中修改theme,supportsRtl.t ...