依据题意可构造出方程组。方程组的每一个方程格式均为：C1*x1 + C2*x2 + ...... + C9*x9 = sum + 4*ki;

高斯消元构造上三角矩阵，以最后一个一行为例：

C*x9 = sum + 4*k。exgcd求出符合范围的x9，其它方程在代入已知的变量后格式亦如此。

第一发Gauss。蛮激动的。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <ctime>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-8)
#define LL long long
#define ULL unsigned long long
#define _LL __int64
#define INF 0x3f3f3f3f
#define Mod 6000007

using namespace std;

const int MAXN = 20;

int up[] = {0,4,3,4,3,5,3,4,3,4};

int site[10][5] = {
{0},
{1,2,4,5},
{1,2,3},
{2,3,5,6},
{1,4,7},
{2,4,5,6,8},
{3,6,9},
{4,5,7,8},
{7,8,9},
{5,6,8,9}
};

int Map[10];

LL coe[MAXN][MAXN];
LL sol[MAXN];

void Output()
{
    int i,j;
    for(i = 1;i <= 9; ++i)
    {
        for(j = 1;j <= 10; ++j)
        {
            printf("%lld ",coe[i][j]);
            if(j == 9)
                printf("= ");
        }
        printf("\n");
    }
    puts("");
}

LL Abs(LL x)
{
    if(x < 0)
        return -x;
    return x;
}

LL gcd(LL x,LL y)
{
    if(y == 0)
        return x;
    return gcd(y,x%y);
}

void exgcd(LL a,LL b,LL &x,LL &y)
{
    if(b == 0)
        x = 1,y = 0;
    else
    {
        LL x1,y1;
        exgcd(b,a%b,x1,y1);
        x = y1;
        y = x1-a/b*y1;
    }
}

//n为行数，m为列数(包括最后一项)
//return -1无整数解 return 0存在整数解。
int Gauss(int n,int m)
{
    int i,j,k;

    LL T,A,B;

    //Output();

    for(i = 1;i < n; ++i)
    {
        for(j = i+1;j <= n; ++j)
        {
            if(coe[j][i] == 0)
                continue;

            if(coe[i][i] == 0)
            {
                for(k = i;k <= m; ++k)
                    T = coe[i][k],coe[i][k] = coe[j][k],coe[j][k] = T;
                continue;
            }

            T = gcd(coe[i][i],coe[j][i]);
            A = coe[j][i]/T,B = coe[i][i]/T;

            for(k = i;k <= m; ++k)
                coe[j][k] = coe[i][k]*A - coe[j][k]*B;
        }
        //Output();
    }

    LL sum = 0;

    for(i = n;i >= 1; --i)
    {
        sum = coe[i][m];
        for(j = m-1;j > i; --j)
            sum -= coe[i][j]*sol[j];

        LL A = coe[i][i],B = 4,C = sum;
        LL x,y;

        exgcd(A,B,x,y);
        //cout<<"A = "<<A<<" B = "<<B<<" C = "<<C<<" x = "<<x<<" y = "<<y<<endl;
        x *= C/gcd(A,B);
        //cout<<"x = "<<x<<endl;
        y = B/gcd(A,B);
        x = (x-x/y*y + Abs(y))%Abs(y);
        sol[i] = x;

        //cout<<"i = "<<i<<" x = "<<x<<endl;

//        if(sum%coe[i][i] != 0)
//            return -1;//此时无整数解
//        sol[i] = sum/coe[i][i];
    }

    return 0;
}

int main()
{
    int i,j;

    for(i = 1;i <= 9; ++i)
        scanf("%d",&Map[i]);

    memset(coe,0,sizeof(coe));

    for(i = 1;i <= 9; ++i)
    {
        for(j = 0;j < up[i]; ++j)
        {
            coe[site[i][j]][i] = 1;
        }
    }

    for(i = 1;i <= 9; ++i)
        coe[i][10] = (4-Map[i])%4;

    if(-1 == Gauss(9,10))
        while(0)
        ;

    bool mark = true;

    for(i = 1;i <= 9;++i)
    {
        for(j = 0;j < sol[i]; ++j)
        {
            if(mark == false)
                printf(" ");
            else
                mark = false;
            printf("%d",i);
        }
    }

    return 0;
}