E. Mishka and Divisors Codeforces Round #365 (Div. 2) 01背包

E. Mishka and Diviso
4000
rs

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

After playing with her beautiful array, Mishka decided to learn some math. After learning how to multiply, divide and what is divisibility, she is now interested in solving the following problem.

You are given integer k and array
a1, a2, ..., an ofn integers. You are to find
non-empty subsequence of array elements such that the product of its elements is divisible byk and it contains minimum possible number of elements.

Formally, you are to find a sequence of indices 1 ≤ i1 < i2 < ... < im ≤ n such that


is divisible byk while
m is minimum possible among all such variants.

If there are more than one such subsequences, you should choose one among them, such that sum of its elements isminimum possible.

Mishka quickly solved this problem. Will you do so?

Input
The first line of the input contains two integers n andk (1 ≤ n ≤ 1 000,1 ≤ k ≤ 1012).

The second line of the input contains n integersa1, a2, ..., an
(1 ≤ ai ≤ 1012) — array elements.

Output
Print single positive integer m in the first line — the number of elements in desired sequence.

In the second line print m distinct integers — the sequence of indices of given array elements, which should be taken into the desired sequence.

If there are more than one such subsequence (e.g. subsequence of minimum possible number of elements and with minimum possible sum of elements), you can print any of them.

If there are no such subsequences, print  - 1 in the only line.

Example

Input

5 60
2 4 6 5 2

Output

3
4 3 1

做这种难题（对于我）真的爽，就算开始一点头绪也没有，但是如果弄懂了会学到很多东西。
这道题０１背包，看题目描述的话应该想到类似的解法，就是每个数字只能选一次，让你求出选最优状态，这就类似０１背包。
那么怎么规范状态呢。
我们分析一下就可以知道，比如说６０的因数是１　２　３　４　５　６　１０　１２　１５　２０　３０　６０这１２个数。
设ｄｐ【ｉ】【ｊ】的意义为前ｉ个数中选最少的数字相乘为ｊ的倍数的数量。
１．ｄｐ[i][j]非递减
２．我们想要得到６０，我们可以从剩余１１个数字状态转移而来、
３．我们结合０１背包，每个数字只有取或者不取，如果不取，那么ｄｐ【ｉ】【ｊ】＝ｄｐ【ｉ－１】【ｊ】，去了就是ｄｐ【ｉ】【ｊ】＝ｄｐ【ｉ－１】【ｊ／ｇｃｄ（ｊ，ｓ【ｉ】）】
＋１（ｓ【ｉ】为给定的数字）
为什么要从ｄｐ【ｉ－１】【ｊ／ｇｃｄ（ｊ，ｓ【ｉ】）】转移而来呢？刚刚说过，ｄｐ数组非递减，那么我们要使ｄｐ【ｉ】【ｊ】中的ｊ尽可能的小，这样得出的结果才能尽可能的小。
如果ｊ／ｇｃｄ（ｊ，ｓ【ｉ】）都要比不放要大的话，那么其他的因数自然也就不成立，如果无法达成ｊ／ｇｃｄ（ｊ，ｓ【ｉ】）这个状态的话，那么其他的因数状态也无法完成，所以这就是正确
的转移方式。
注意，这道题卡常数，非常恶心。在ｇｃｄ的时候，不要直接ｇｃｄ（ｊ，ｓ【ｉ】），要先ｓｔ【ｉ】＝ｇｃｄ（ｋ，ｓ【ｉ】），然后ｊ／ｇｃｄ（ｊ，ｓｔ【ｉ】），这样可以少跑一点。

#include<bits/stdc++.h>
using namespace std;
long long s[1008],st[1008];
int pre[1008][10080];
int dp[10008];
long long sum[10008];
vector<long long> factor;
map<long long,int>maped;
int div(long long k)
{
for(long long i=1; i*i<=k; i++)
{
if(k%i==0)
{
factor.push_back(i);
factor.push_back(k/i);
}
}
sort(factor.begin(),factor.end());
return unique(factor.begin(),factor.end())-factor.begin();
}
int main()
{
int n;
long long k;
long long Min=LLONG_MAX;
int pos;
while(scanf("%d%I64d",&n,&k)!=EOF)
{
for(int i=1; i<=n; i++)
{
scanf("%I64d",&s[i]);
st[i]=__gcd(k,s[i]);
if(Min>s[i])
{
Min=s[i];
pos=i;
}
}
if(k==1)
{
printf("%d\n",1);
printf("%d\n",pos);
continue;
}
memset(pre,-1,sizeof(pre));
memset(sum,0x3f,sizeof(sum));
memset(dp,0x3f,sizeof(dp));
dp[0]=sum[0]=0;
factor.clear();
maped.clear();
int len=div(k);
for(int i=0; i<len; i++)
{
maped[factor[i]]=i;
}
for(int i=1; i<=n; i++)
{
for(int j=len-1; j>=1; j--)
{
long long t=__gcd(factor[j],(long long)st[i]);
int pos=maped[factor[j]/t];
if(dp[j]>dp[pos]+1)
{
dp[j]=dp[pos]+1;
sum[j]=sum[pos]+s[i];
pre[i][j]=pos;
}
if(dp[j]==dp[pos]+1&&sum[j]>sum[pos]+s[i])
{
sum[j]=sum[pos]+s[i];
pre[i][j]=pos;
}
}
}
if(dp[len-1]>n)
{
printf("-1\n");
}
else
{
printf("%d\n",dp[len-1]);
int a=n,b=len-1;
while(a>0&&b>0)
{
while(pre[a][b]==-1&&a>0&&b>0)
a--;
printf("%d ",a);
b=pre[a--][b];
}

}
printf("\n");

}
}