"Problem solution": a sequence

. A: sequence d

Time limit: 1 Sec   Memory Limit: 512 MB

Face questions

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He declined to publicly face the question.

answer

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Simple configuration problem. Think positive solutions (no brainer good cutting) However deceived points blinded when the game ......

In fact after the game to change the topic, I found that my thinking is too simple, even if can not afford to think positive solution also cut during the Games.

inf decision: The hands may find this play four.

 

if(a+b>n+1){puts("No");continue;}
if(a*b<n){puts("No");continue;}
if(a>n||b>n){puts("No");continue;}
if(a==0||b==0){puts("No");continue;}

 

For the data of $ 20% $, $ n \ leq 5 $, $ next-permutation $ directly to violence.

$ $ For the data of 30%, to ensure $ N = A * B $, is the real part of this solution to give positive tips brought portion. Play with the hand can be found, we can $ N $ elements $ B $ bisected into groups, within each group monotonically increasing, monotonically decreasing between the groups, must be constructed so that a legitimate program.

For data guaranteed $ 20% $ $ B \ leq 2 $, so that the case can be divided directly play hand. Where $ B = 1 $ is very simple, if there is a legitimate solution at this time, must satisfy $ A = n $. $ B = 2 $ circumstances of the way I write is very sb not elaborate. Details look at the code.

Data for $ 100% $ of positive solutions from the above $ 30% $ data from the promotion. We still $ B $ can be divided into groups, the groups monotonously increasing monotonically decreased groups, then we need to ensure that only the largest group of elements in a $ A $, and the number of elements remaining group not exceeded $ A $ It can be.

This is still a good implementation. But I thought too rigid, has been using a number of methods weird, try using the four operations $ O (1) $ obtained a legal decomposition scheme. However, not so easy to find. Finally opened the array to maintain the size of each group.

Code:

 

#include<bits/stdc++.h>
#define lowbit(A) A&-A
#define rint register int
using namespace std;
int T,n,a,b,pl[100005],t[15],dp[15],up,dn,tim,len[100005];
bool vis[100005],is_ok;
inline void update(int x,int dat)
{
    for(;x<=n;x+=lowbit(x))
        t[x]=max(t[x],dat);
    return ;
}
inline int query(int x)
{
    int res=0;
    for(;x;x-=lowbit(x))
        res=max(res,t[x]);
    return res;
}
int main ()
{
    scanf("%d",&T);
    while(T--)
    {
        is_ok=0;
        scanf("%d %d %d",&n,&a,&b);
        if(a+b>n+1){puts("No");continue;}
        if(a*b<n){puts("No");continue;}
        if(a>n||b>n){puts("No");continue;}
        if(a==0||b==0){puts("No");continue;}
        if(n<=5)
        {
            for(rint i=1;i<=n;++i)pl[i]=i;
            do{
                for(rint i=1;i<=n;++i)t[i]=0,dp[i]=1;up=1;
                for(rint i=1;i<=n;++i)
                {
                    dp[i]=query(pl[i]-1)+1;
                    up=max(up,dp[i]);
                    update(pl[i],dp[i]);
                }
                if(up!=a)continue;
                for(rint i=1;i<=n;++i)t[i]=0,dp[i]=1;dn=1;
                for(rint i=n;i>=1;--i)
                {
                    dp[i]=query(pl[i]-1)+1;
                    dn=max(dn,dp[i]);
                    update(pl[i],dp[i]);
                }
                if(dn==b){is_ok=1;break;}
            }while(next_permutation(pl+1,pl+n+1));
            if(!is_ok)puts("No");
            else
            {
                puts("Yes");
                for(rint i=1;i<=n;++i)
                    printf("%d ",pl[i]);
                puts("");
            }
            continue;
        }
        else if(a*b==n)
        {
            puts("Yes");
            for(rint i=1;i<=a;++i)
            {
                for(rint j=b*i;j>b*(i-1);--j)
                    printf("%d ",j);
            }
            puts("");
            continue;
        }
        else if(b==1)
        {
            puts("Yes");
            for(rint i=1;i<=n;++i)printf("%d ",i);puts("");
            continue;
        }
        else if(b==2)
        {
            int lin=n/2;
            int bro=a-lin-(n&1);puts("Yes");
            for(rint i=1;i<=2*bro;++i)
                printf("%d ",i);
            for(rint i=bro+1;i<=lin;++i)
                printf("%d %d ",2*i,2*i-1);
            if(n&1)printf("%d",n);
            puts("");
            continue;
        }
        puts("Yes");
        int zh=n;
        memset(len,0,sizeof(len));
        for(rint j=zh-a+1;j<=zh;++j)printf("%d ",j);zh=n-a;
        for(rint i=1;i<=n-a;++i)
            len[i%(b-1)+1]++;
        for(rint i=1;i<=b-1;++i)
        {
            for(rint j=zh-len[i]+1;j<=zh;++j)printf("%d ",j);
            zh = zh- len [i];
        }
        puts("");
    }
    return 0;
}

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