Codeforces 740A Alyona and copybooks

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本题探讨了如何以最少花费购买字帖使总数为4的倍数的问题。女孩Alyona需要为四门课程准备等量字帖，面对三种不同包装的价格，找出最优购买策略。

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.

   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：

需要的字帖数目需要是 4 的整数倍，现在已经拥有 n 本字帖，有三种选购方案，输出最小花费。

思路：

单独讨论还缺少0-3本的情况，注意，并不是递增的价格！即缺少一本的时候，可以买一本，也可以买五本，只要省钱。

AC CODE：

#include<stdio.h>
#include<cstring>
#include<algorithm>
#define HardBoy main()
#define ForMyLove return 0;
using namespace std;
typedef long long LL;
const int MYDD = 1103;

int HardBoy {
	LL n, a, b, c;
	while(scanf("%lld %lld %lld %lld", &n, &a, &b, &c) != EOF) {
		LL ans;
		LL s = 4 - n%4;
		if(s == 4) {
			ans = 0;
		} else if(s == 3) {
			ans = min(a+b, min(3*a, c));
		} else if(s == 2) {
			ans = min(min(2*a, b), 2*c);
		} else if(s == 1) {
			ans = min(min(a, 3*c), b+c);
		}
		printf("%lld\n", ans);
	}
	ForMyLove
}

思维不成熟的 CODE：

#include<stdio.h>
#include<cstring>
#include<algorithm>
#define HardBoy main()
#define ForMyLove return 0;
using namespace std;
const int MYDD = 1103;

int HardBoy {
	int n, a, b, c;
	while(scanf("%d %d %d %d", &n, &a, &b, &c) != EOF) {
		int ans;
		int s = 4 - n%4;
		if(s == 4) {
			ans = 0;
		} else if(s == 3) {
			ans = min(a+b, min(3*a, c));
		} else if(s == 2) {
			ans = min(2*a, b);
		} else if(s == 1) {
			ans = a;
		}
		printf("%d\n", ans);
	}
	ForMyLove
}