博客
关于我
A. Cheap Travel
阅读量:533 次
发布时间:2019-03-09

本文共 1453 字,大约阅读时间需要 4 分钟。

Ann has recently started commuting by subway. We know that a one ride subway ticket costs a rubles. Besides, Ann found out that she can buy a special ticket for m rides (she can buy it several times). It costs b rubles. Ann did the math; she will need to use subway n times. Help Ann, tell her what is the minimum sum of money she will have to spend to make n rides?

Input

The single line contains four space-separated integers n, m, a, b (1 ≤ n, m, a, b ≤ 1000) — the number of rides Ann has planned, the number of rides covered by the m ride ticket, the price of a one ride ticket and the price of an m ride ticket.

Output

Print a single integer — the minimum sum in rubles that Ann will need to spend.

Examples

inputCopy
6 2 1 2
outputCopy
6
inputCopy
5 2 2 3
outputCopy
8
Note
In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three m ride tickets.

思路:模拟

写的时候被三组数据卡了,批评自己
10 3 5 1
101 110 1 100
100 8 10 1

#include
#include
#include
#include
#include
#include
//#include
using namespace std;typedef long long LL;int main(){ LL n, m, a, b; cin >> n >> m >> a >> b; if (n / m == 0) { if (b < a*n) cout << b << endl; else cout << a * n << endl; } else if (b*1.0 / m >= a) cout << n * a << endl; else { if(b<(n%m)*a) cout << n / m * b + b << endl; else cout << n / m * b + (n%m)*a << endl; } return 0;}

转载地址:http://vboiz.baihongyu.com/

你可能感兴趣的文章
mysql-connector-java各种版本下载地址
查看>>
mysql-EXPLAIN
查看>>
MySQL-Explain的详解
查看>>
mysql-group_concat
查看>>
MySQL-redo日志
查看>>
MySQL-【1】配置
查看>>
MySQL-【4】基本操作
查看>>
Mysql-丢失更新
查看>>
Mysql-事务阻塞
查看>>
Mysql-存储引擎
查看>>
mysql-开启慢查询&所有操作记录日志
查看>>
MySQL-数据目录
查看>>
MySQL-数据页的结构
查看>>
MySQL-架构篇
查看>>
MySQL-索引的分类(聚簇索引、二级索引、联合索引)
查看>>
Mysql-触发器及创建触发器失败原因
查看>>
MySQL-连接
查看>>
mysql-递归查询(二)
查看>>
MySQL5.1安装
查看>>
mysql5.5和5.6版本间的坑
查看>>