Educational Codeforces Round 26 D Round Subset

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D. Round Subset

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Let's call the roundness of the number the number of zeros to which it ends.

You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum
possible.

Input

The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).

The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018).

Output

Print maximal roundness of product of the chosen subset of length k.

Examples

input

3 2
50 4 20

output

3

input

5 315 16 3 25 9

output

3

input

3 39 77 13

output

0

Note

In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] —
product 80, roundness 1,[50, 20] — product 1000, roundness 3.

In the second example subset [15, 16, 25] has product 6000, roundness 3.

In the third example all subsets has product with roundness 0.

题意是给定n个数，从中选取k个数，问这k个数相等得到的数中，最多有几个0.

都结束了才做的题，发现标签是DP，不然怎么也想不到是DP。离散中学到0是由2和5相乘得到的。然后二维DP，(i,j)表示第i个数取前j个最大的可能，推不出递推式，一狠心设了个4维DP，后两位表示2和5的个数，又觉得i，j没用了。想了好久，觉得用i表示前n个数，j表示前i个数中所有质因子为2的和，dp[i][j]表示当前状态下所有质因子为5的个数的和。这样只需要数组的第二维开的稍大些，就可以解决。然后可以得到递推式：dp[i][j]=max(dp[i][j],dp[i-1][j-ans2]+ans5);最后确定了k的个数，扫描j，j从i到数组结束，注意要取2和5个数小的那一个，所以是min(i,dp[k][i])。注意long
long的使用，防止爆数据。

代码实现：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#define ll long long

using namespace std;
const int maxn=205;
ll dp[maxn][60*maxn];
ll map[maxn],ans2;

ll judge(ll x)
{
ll ans1=0;
ll temp=x;
ans2=0;
while(temp)
{
if(x%2==0)
{
ans1++;
x/=2;
}
else
break;
}
while(temp)
{
if(x%5==0)
{
ans2++;
x/=5;
}
else
break;
}

return ans1;
}

int main()
{
ll n,k,i,j,x,ans1,l;
while(cin>>n>>k)
{
memset(dp,-0x3f3f3f3f,sizeof(dp));
dp[0][0]=0;

for(i=0;i<n;i++)
{
cin>>x;
ans1=judge(x);
for(j=k;j>=1;j--)
{
for(l=ans1;l<60*maxn;l++)
{
dp[j][l]=max(dp[j][l],dp[j-1][l-ans1]+ans2);
}
}
}
ll ans=0;
for(i=1;i<60*maxn;i++)
{
ans=max(ans,min(i,dp[k][i]));
}
cout<<ans<<endl;
}
return 0;
}