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Codeforces 703E

2016-12-08 10:59 309 查看
num[i][j],sum[i][j],前i个数合成k的第j的因子至少需要个数以及最小和

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef __int64 LL;
const LL INF=1e17;
const int N=1000;
LL aa
,cc
,bb[N+10000];int top;
int num
[N+10000];LL sum
[N+10000];int pre
[N+10000];

LL gcd(LL a,LL b){
if(!b)return a;
return gcd(b,a%b);
}
int id(LL a){
return lower_bound(bb,bb+top,a)-bb;
}

int main(){
#ifdef DouBi
freopen("in.cpp","r",stdin);
#endif // DouBi
int n;LL k;
while(scanf("%d%I64d",&n,&k)!=EOF){
LL tk=k;
for(int i=0;i<n;i++){
scanf("%I64d",&aa[i]);
cc[i]=gcd(aa[i],k);
tk/=gcd(tk,cc[i]);
}
if(tk!=1){
printf("-1\n");continue;
}
//printf("in1\n");
top=0;
for(int j=1;j<=k/j;j++){
if(k%j==0)bb[top++]=j;
}
//printf("in2\n");
if(k/bb[top-1]!=bb[top-1]){
bb[top]=k/bb[top-1];top++;
for(int i=top-3;i>=0;i--)bb[top++]=k/bb[i];
}
else {
for(int i=top-2;i>=0;i--)bb[top++]=k/bb[i];
}

//printf("in3\n");
//for(int j=0;j<top;j++)printf("%I64d ",bb[j]);printf("bb\n");

for(int j=0;j<top;j++){
if(aa[0]%bb[j]==0)sum[0][j]=aa[0],num[0][j]=1,pre[0][j]=-1;
else sum[0][j]=INF,num[0][j]=n+10,pre[0][j]=-1;
}
for(int i=1;i<n;i++){
for(int j=0;j<top;j++){
num[i][j]=num[i-1][j];sum[i][j]=sum[i-1][j];pre[i][j]=j;
//dp[i][j]=min(dp[i-1][j],dp[i-1][bb[j]/gcd(bb[j],aa[i])]+1);
int idx=id(bb[j]/gcd(bb[j],cc[i]));
int x=num[i-1][idx]+1;
LL y=sum[i-1][idx]+aa[i];
if(idx==0){
x=1;
y=aa[i];
}
if(x<num[i][j]||x==num[i][j]&&y<sum[i][j]){
num[i][j]=x;sum[i][j]=y;
if(idx==0)idx=-1;
pre[i][j]=idx;
}
if(sum[i][j]>INF)sum[i][j]=INF;
}
}

printf("%d\n",num[n-1][top-1]);
int j=top-1;
for(int i=n-1;i>=0;i--){
if(j==-1)break;
if(pre[i][j]!=j){
if(j==top-1)printf("%d",i+1);
else printf(" %d",i+1);
}
j=pre[i][j];
}
printf("\n");
}
return 0;
}
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