CodeForces 740A - Alyona and copybooks(思维)

A. Alyona and copybooks
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook’s packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.

What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.

Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).

Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.

Examples
input
1 1 3 4
output
3
input
6 2 1 1
output
1
input
4 4 4 4
output
0
input
999999999 1000000000 1000000000 1000000000
output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.

In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.

In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn’t need to buy anything.

In the fourth example Alyona should buy one pack of one copybook.

题意：
已经有了n个本子，要再买若干本子，使得本子总数是4的倍数。

花a卢布能买1本，b卢布能买2本，c卢布能买三本。

问所花费的最少的钱数。

解题思路：
暴力搜索。

AC代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long n,a,b,c;
    cin>>n>>a>>b>>c;
    if(n%4==0)  printf("0");
    else
    {
        long long minn = LONG_LONG_MAX;
        for(int i = 0;i <= 4;i++)
            for(int j = 0;j <= 4;j++)
                for(int k = 0;k <= 4;k++)
                    if((n+i+j*2+k*3)%4 == 0)    minn = min(minn,i*a+j*b+k*c);
        cout<<minn;
    }
    return 0;
}