Description
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 arubles, 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.
Sample Input
1 1 3 4
3
6 2 1 1
1
4 4 4 4
0
999999999 1000000000 1000000000 1000000000
1000000000
Hint
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.
不是太难，就是考虑的情况多了点，
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
ll n,a,b,c;
int main()
{
cin >> n>> a >> b>> c;
if(n%4==0)
{
cout << "0" <<endl;
return 0;
}
ll k = 4 - n%4;
if(k==1)
{
ll t1 = b+c; //考虑2+3
t1 = min(t1,3*c);//考虑1+1+1
if(a<t1)
{
cout << a <<endl;
}
else
cout << t1<< endl;
}
else if(k==2)
{
ll t1 = 0;
if(c>a*3)
t1 = a*6;//1+1+1+1+1+1
else
t1 = c*2;//3+3
t1 = min(t1,a*2);//1+1
if(t1 < b)
cout << t1 <<endl;
else
cout << b <<endl;
}
else if(k==3)
{
ll t1 = b+a;//1+2
t1 = min(t1,a*3);//1+1+1
if(t1 < c)
cout << t1 <<endl;
else
cout << c <<endl;
}
return 0;
}