time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Recently Anton found a box with digits in his room. There are k2 digits 2, k3 digits 3, k5 digits 5 and k6 digits 6.
Anton’s favorite integers are 32 and 256. He decided to compose this integers from digits he has. He wants to make the sum of these integers as large as possible. Help him solve this task!
Each digit can be used no more than once, i.e. the composed integers should contain no more than k2 digits 2, k3 digits 3 and so on. Of course, unused digits are not counted in the sum.
Input
The only line of the input contains four integers k2, k3, k5 and k6 — the number of digits 2, 3, 5 and 6 respectively (0 ≤ k2, k3, k5, k6 ≤ 5·106).
Output
Print one integer — maximum possible sum of Anton’s favorite integers that can be composed using digits from the box.
Examples
input
5 1 3 4
output
800
input
1 1 1 1
output
256
Note
In the first sample, there are five digits 2, one digit 3, three digits 5 and four digits 6. Anton can compose three integers 256 and one integer 32 to achieve the value 256 + 256 + 256 + 32 = 800. Note, that there is one unused integer 2 and one unused integer 6. They are not counted in the answer.
In the second sample, the optimal answer is to create on integer 256, thus the answer is 256.
【题目链接】:http://codeforces.com/contest/734/problem/B
【题解】
贪心。
先选2,5,6这3个数字；
个数为min(k2,k5,k6);
然后再选3,2这两个数字；
个数同样为min(k3,k2);
然后加起来;
不知道会不会超long long;反正开就是了；
【完整代码】
#include <bits/stdc++.h>
#define LL long long
using namespace std;
LL k2,k3,k5,k6;
int main()
{
cin >> k2 >> k3 >>k5 >>k6;
LL ans = 0;
LL temp = min(k2,min(k5,k6));
k2-=temp,k5-=temp,k6-=temp;
ans += temp * 256;
temp = min(k3,k2);
ans += temp * 32;
cout << ans << endl;
return 0;
}