Codeforces#381(Div. 2) A.Alyona and copybooks【暴力】
2016-11-24 13:01
459 查看
A. Alyona and copybooks
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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
Output
Input
Output
Input
Output
Input
Output
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.
题目大意:
我们现在已经有了n个物品,商店卖这种物品有三种方式:
①卖一个,价格为a。
②卖两个,价格为b。
③卖三个,价格为c。
现在要求购买任意方案的数量的该物品,但是需要使得购买量+n是4的倍数。问能够达到目的的最小花费。
思路:
直接暴力枚举每种购买方案的数量即可,比赛中为了稳妥,竟然没有思考的直接裸枚举到了100...............23333333333333333333
其实枚举到4就够了。
Ac代码:
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll __int64
int main()
{
ll n,a,b,c;
while(~scanf("%I64d%I64d%I64d%I64d",&n,&a,&b,&c))
{
if(n%4==0)
{
printf("0\n");
continue;
}
else
{
ll output=2000000000000000000;
for(ll i=0;i<=100;i++)
{
for(ll j=0;j<=100;j++)
{
for(ll k=0;k<=100;k++)
{
if(((i+j*2+k*3)+n)%4==0)
{
output=min(output,a*i+b*j+c*k);
}
}
}
}
printf("%I64d\n",output);
}
}
}
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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.
题目大意:
我们现在已经有了n个物品,商店卖这种物品有三种方式:
①卖一个,价格为a。
②卖两个,价格为b。
③卖三个,价格为c。
现在要求购买任意方案的数量的该物品,但是需要使得购买量+n是4的倍数。问能够达到目的的最小花费。
思路:
直接暴力枚举每种购买方案的数量即可,比赛中为了稳妥,竟然没有思考的直接裸枚举到了100...............23333333333333333333
其实枚举到4就够了。
Ac代码:
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll __int64
int main()
{
ll n,a,b,c;
while(~scanf("%I64d%I64d%I64d%I64d",&n,&a,&b,&c))
{
if(n%4==0)
{
printf("0\n");
continue;
}
else
{
ll output=2000000000000000000;
for(ll i=0;i<=100;i++)
{
for(ll j=0;j<=100;j++)
{
for(ll k=0;k<=100;k++)
{
if(((i+j*2+k*3)+n)%4==0)
{
output=min(output,a*i+b*j+c*k);
}
}
}
}
printf("%I64d\n",output);
}
}
}
相关文章推荐
- Codeforces Round #381 (Div. 2) A. Alyona and copybooks
- Codeforces Round #381 (Div. 2) A. Alyona and copybooks(思维)
- Codeforces#381(Div. 2) B.Alyona and flowers【思维+暴力】
- Codeforces Round #381 (Div. 2) A. Alyona and copybooks
- Codeforces Round #381 (Div. 2) A. Alyona and copybooks(模拟)
- Codeforces Round #381 (Div. 2) A.Alyona and copybooks
- Codeforces Round #381 (Div. 2) A Alyona and copybooks
- Codeforces Round #401 (Div. 2) C. Alyona and Spreadsheet(暴力)
- Alyona and copybooks
- 【20.23%】【codeforces 740A】Alyona and copybooks
- codeforces 381A. Alyona and copybooks=
- Codeforces Round #358 (Div. 2) -- A. Alyona and Numbers(暴力水题)
- Codeforces 740A Alyona and copybooks
- CodeForces 740A Alyona and copybooks
- codeforces 740A Alyona and copybooks
- Codeforces Round #401 (Div. 2)C. Alyona and Spreadsheet(暴暴暴力)
- A. Alyona and copybooks
- A. Alyona and copybooks--思维水题
- Codeforces#381(Div. 2) C. Alyona and mex【思维】好题~
- acm Alyona and copybooks 新人水题