链接：https://www.nowcoder.com/acm/contest/143/J
来源：牛客网
There are n students going to travel. And hotel has two types room:double room and triple room. The price of a double room is p2 and the price of a triple room is p3
Now you need to calulate the minimum total cost of these students.
输入描述:
The first line has three integers n, p2, p3
输出描述:
Output the minimum total cost.
备注:
1<=n<=10^9
1<=p2,p3<=10^9
题意：有n个学生要住酒店，酒店有双人房和三人房，两种房间的价格为p2，p3，问怎么住可以花费最小
分析：所有情况有五种：
所有学生全部住双人间
所有学生全部住三人间
学生先住三人间，如果有剩一到两个人则去住双人间
学生先住三人间，如果有剩四个人则去住两间双人房
学生先住两人间，最后剩一到三人则去住三人间
计算所有情况花费的费用，取最小值
AC代码：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5 + 10;
const ll mod = 1e9 + 7;
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
ll n, p2, p3;
while( cin >> n >> p2 >> p3 ) {
ll ans = (n+1)/2*p2;
ans = min( ans, (n+2)/3*p3 );
ans = min( ans, n/3*p3+p2 );
ans = min( ans, (n-2)/3*p3+p2+p2);
ans = min( ans, (n-2)/2*p2+p3 );
cout << ans << endl;
}
return 0;
}