我的代码不一定是最好的，但一定是最容易看懂的 :) 各位，给个赞同吧

#include <stdio.h>
char verity(char data[81]){
    long right=0,right2=0;
    //验证横竖
    for(int i=0;i<9;i++){
        for(int j=0;j<9;j++){
            //i,j在此处既表示横行，又表示竖列
            right+=1<<data[i*9+j];
            right2+=1<<data[j*9+i];
        }
        //1<<1+1<<2+1<<3+...+1<<9=(1<<10)-2=1022.
        if(right!=(1<<10)-2||right2!=(1<<10)-2)
            return 0;
        right=right2=0;
    }
    right=0;
    //验证九宫格
    for(int a=0;a<3;a++){
        for(int b=0;b<3;b++){
            //a,b表示第a行第b列个九宫格
            //i,j是每个九宫格内的行列数
            for(int i=a*3;i<a*3+3;i++){
                for(int j=b*3;j<b*3+3;j++){
                    right+=1<<data[i*9+j];
                }
                
            }
            if(right!=(1<<10)-2)
                    return 0;
                right=0;
        }
    }
    return 1;

}
int main(int argc, char* argv[])
{
    int N;
    scanf("%d",&N);
    //不用[N]，我在本地调试时用的VC++不支持C99标准
    char data[20][81];
    //输入数据
    for(int j=0;j<N;j++){
        for(int i=0;i<81;i++){
            scanf("%d",&data[j][i]);
        }
        //验证
        if(verity(data[j]))
            printf("Right");
        else
            printf("Wrong");
        printf("\n");
    }
    return 0;
}

模拟

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int ALL=(1<<9)-1;
int mp[9][9];
bool chk_l(int x){
    int tmp=0;
    for ( int i=0; i<9; i++ ) tmp |= (1<<mp[x][i]);
    return tmp==ALL;
}
bool chk_r(int x){
    int tmp=0;
    for ( int i=0; i<9; i++ ) tmp |= (1<<mp[i][x]);
    return tmp==ALL;
}
bool chk_g(int x,int y){
    int tmp=0;
    for ( int i=x; i<=x+2; i++ )
        for ( int j=y; j<=y+2; j++ )
            tmp |= (1<<mp[i][j]);
    return tmp==ALL; 
}
bool chk(){
    for ( int i=0; i<9; i++ ) if( !chk_l(i) ) return false;
    for ( int i=0; i<9; i++ ) if( !chk_r(i) ) return false;
    for ( int i=0; i<9; i+=3)
        for ( int j=0; j<9; j+=3)
            if( !chk_g(i,j) ) return false;
    return true; 
}
int main(){
    int T;
    scanf("%d", &T);
    while( T-- ){
        for ( int i=0; i<9; i++ )
            for ( int j=0; j<9; j++ )
                scanf("%d", &mp[i][j] ), mp[i][j]--;
        chk()?puts("Right"):puts("Wrong");
    } 
    return 0;
}

横向、竖向和九宫格是可以同时在一个循环里面检测的，只要有一个不满足即可以结束检测，因此可以极大的减少代码量

#include <iostream>
#include <cstring>

using namespace std;

int arr[9][9];

int main()
{
    //freopen("input.txt", "r", stdin);
    
    int i, j, k;
    int n;
    scanf("%d", &n);
    
    int arr_idx[3][9];
    while(n--)
    {
        for(i = 0;i < 9; ++i)
        {
            for(j = 0; j < 9; ++j)
            {
                scanf("%d", &arr[i][j]);
            }
        }
        
        //横竖向检测 
        int mapi[9] = {0, 0, 0, 3, 3, 3, 6, 6, 6};
        int mapj[9] = {0, 3, 6, 0, 3, 6, 0, 3, 6};
        
        for(i = 0; i < 9; ++i)
        {
            memset(arr_idx, 0, sizeof(arr_idx));

            for(j = 0; j < 9; ++j)
            {
                if(++arr_idx[0][arr[i][j] - 1] > 1) break; //横向 
                if(++arr_idx[1][arr[j][i] - 1] > 1) break; //竖向 
                if(++arr_idx[2][arr[j / 3 + mapi[i]][j % 3 + mapj[i]] - 1] > 1) break; //九宫格 
            }
            if(j < 9) break;
        }
        if(i < 9)
        {
            printf("Wrong\n");
        }
        else
        {
            printf("Right\n");
        }    
    }
    
    //fclose(stdin);
    
    return 0;
}

Ha

var
 a:array[1..9,1..9]of longint;
 f:array[1..9]of boolean;
 i,j,k,l,m,n:longint;
 rw:boolean;
begin
 readln(n);
 for i:=1 to n do
  begin
   rw:=true;
   for j:=1 to 9 do
    for k:=1 to 9 do
     read(a[j,k]);
   for j:=1 to 9 do
    begin
     fillchar(f,sizeof(f),false);
     for k:=1 to 9 do
      f[a[j,k]]:=true;
     for k:=1 to 9 do
      if f[k]=false then begin
                          writeln('Wrong');
                          rw:=false;
                          break;
                         end;
     if rw=false then break;
     fillchar(f,sizeof(f),false);
     for k:=1 to 9 do
      f[a[k,j]]:=true;
     for k:=1 to 9 do
      if f[k]=false then begin
                          writeln('Wrong');
                          rw:=false;
                          break;
                         end;
     if rw=false then break;
    end;
   if rw=false then continue;
   for j:=0 to 2 do
    begin
     for k:=0 to 2 do
      begin
       fillchar(f,sizeof(f),false);
       for l:=1 to 3 do
        for m:=1 to 3 do
         f[a[j*3+l,k*3+m]]:=true;
       for l:=1 to 9 do
        if f[l]=false then begin
                            writeln('Wrong');
                            rw:=false;
                            break;
                           end;
       if rw=false then break;
      end;
    if rw=false then break;
   end;
  if rw=false then continue;
  writeln('Right');
 end;
end.

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <algorithm>
using namespace std;

int main(){
    int i,n,a[10][10],j,k;
    scanf("%d",&n);
    for(k=1;k<=n;k++)
    {
        int e=1;
        for(i=1;i<=9;i++)
        {
            for(j=1;j<=9;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        if(k!=n)
            getchar();
        for(i=1;i<=9;i++)
        {
            int b[10]={0};
                b[0]=1;
            for(j=1;j<=9;j++)
            {
                if(b[a[i][j]]==0)
                    b[a[i][j]]=1;
                else
                {
                    printf("Wrong\n");
                    e=0;
                    break;
                }
            }
            if(e==0)
                break;
        }
        if(e==0)
            continue;
        for(j=1;j<=9;j++)
        {
            int b[10]={0};
                b[0]=1;
            for(i=1;i<=9;i++)
            {
                if(b[a[i][j]]==0)
                    b[a[i][j]]=1;
                else
                {
                    printf("Wrong\n");
                    e=0;
                    break;
                }
            }
            if(e==0)
                break;
        }
        if(e==0)
            continue;
        for(i=2;i<=8;i=i+3)
        {
            for(j=2;j<=8;j=j+3)
            {
                int b[10]={0};
                b[0]=1;
                if(b[a[i][j]]==0)
                    b[a[i][j]]=1;
                else
                    e=0;
                if(b[a[i-1][j-1]]==0)
                    b[a[i-1][j-1]]=1;
                else
                    e=0;
                if(b[a[i][j-1]]==0)
                    b[a[i][j-1]]=1;
                else
                    e=0;
                if(b[a[i-1][j]]==0)
                    b[a[i-1][j]]=1;
                else
                    e=0;
                if(b[a[i+1][j]]==0)
                    b[a[i+1][j]]=1;
                else
                    e=0;
                if(b[a[i][j+1]]==0)
                    b[a[i][j+1]]=1;
                else
                    e=0;
                if(b[a[i+1][j+1]]==0)
                    b[a[i+1][j+1]]=1;
                else
                    e=0;
                if(b[a[i-1][j+1]]==0)
                    b[a[i-1][j+1]]=1;
                else
                    e=0;
                if(b[a[i+1][j-1]]==0)
                    b[a[i+1][j-1]]=1;
                else
                    e=0;
                if(e==0)
                    break;
            }
            if(e==0)
            break;
        }
        if(e!=0)
            printf("Right\n");
        else
            printf("Wrong\n");
    }
    return 0;
}

毫不夸张的说，我的代码思路清晰，也容易看懂（小骄傲）
如果觉得好的话，点个赞吧。
思路
1.用#三维数组#实现多个数独的输入
2.定义判断数独是否正确的函数Sudoku
3.在这个判断函数中，要对 行 列 3×3矩阵 进行判断
4.定义该判断函数Jcomplete判断1-9的的完整性
下面给出代码：

#include<iostream>
using namespace std;
bool Jcomplete(const int x[]) {//1-9判断
    for (int i = 1; i < 10; i++) {
        int j;
        for (j = 0; j < 9; j++)if (x[j] == i)break;
        if (j == 9)return false;
    }
    return true;
}
bool Sudoku(int x[][9]) {//数独判断
    int d = 1, y[9];
    for (int i = 0; i < 9; i++) {
        if (!Jcomplete(x[i]))return false;//行判断
        for (int j = 0; j < 9; j++)y[j] = x[i][j];
        if (!Jcomplete(y))return false;//列判断
        for (int j = 0; j < 9; j++)y[j] = x[i / 3 * 3 + j / 3][3 * i % 9 + j % 3];
        if (!Jcomplete(y))return false;//3*3判断
    }
    return true;
}
int main()
{
    int N;
    int x[10][9][9];
    cin >> N;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < 9; j++)
            for (int k = 0; k < 9; k++)
                cin >> x[N][j][k];
        if (Sudoku(x[N]))cout << "Right" << endl;
        else cout << "Wrong" << endl;
    }
    return 0;
}

不用写这么麻烦的1~9判断，只需要将9个数求和等于45，求积等于362880就可以确定是1~9了。

#include <iostream>

using namespace std;

int main()
{
    int cr,cc;
    cin>>cc;
    for(cr=0;cr<cc;cr++)
    {
        bool flag=true;
        int sd[9][9];
        int i,j,k,l;
        int sum;
        long mu;
        for(i=0;i<9;i++)
        {
            for(j=0;j<9;j++)
            {
                cin>>sd[i][j];
            }
        }
        for(i=0;i<9;i++)
        {
            sum=0;
            mu=1;
            for(j=0;j<9;j++)
            {
                sum+=sd[i][j];
                mu*=sd[i][j];
            }
            if(sum!=45||mu!=362880)
            {
                flag=false;
            }
        }
        for(i=0;i<9;i++)
        {
            sum=0;
            mu=1;
            for(j=0;j<9;j++)
            {
                sum+=sd[j][i];
                mu*=sd[j][i];
            }
            if(sum!=45||mu!=362880)
            {
                flag=false;
            }
        }
        for(i=0;i<9;i+=3)
        {
            for(j=0;j<9;j+=3)
            {
                sum=0;
                mu=1;
                for(k=0;k<3;k++)
                {
                    for(l=0;l<3;l++)
                    {
                        sum+=sd[i+k][j+l];
                        mu*=sd[i+k][j+l];
                    }
                }
                if(sum!=45||mu!=362880)
                {
                    flag=false;
                }
            }
        }
        if(flag)
        {
            cout<<"Right"<<endl;
        }
        else
        {
            cout<<"Wrong"<<endl;
        }
    }
    return 0;
}

函数中判断错误可增加输出。
#include <iostream>
using namespace std;

int n;
char a;
int arr[9][9];
int brr[24];

int identify() {
for (int j = 0; j < 9; j++) {
int crr[10];
for(int i=1;i<=9;i++){
crr[i]=0;
}
for (int k = 0; k < 9; k++) {
if (crr[arr[j][k]] == 0) {
crr[arr[j][k]] = 1;
} else {
return 0;
}
}
}
//cout<<1<<endl;
for (int k = 0; k < 9; k++) {
int crr[10];
for(int i=1;i<=9;i++){
crr[i]=0;
}
for (int j = 0; j < 9; j++) {
if (crr[arr[j][k]] == 0) {
crr[arr[j][k]] = 1;
} else {
return 0;
}
}
}
//cout<<2<<endl;
for (int j = 1; j <= 7; j += 3) {
for (int k = 1; k <= 7; k += 3) {
int crr[10];
for(int i=1;i<=9;i++){
crr[i]=0;
}
for (int x = j - 1; x <= j + 1; x++) {
for (int y = j - 1; y <= j + 1; y++) {
if (crr[arr[x][y]] == 0) {
crr[arr[x][y]] = 1;
} else {
return 0;
}
}
}
}
}
return 1;
}

int main() {
cin >> n;
for (int i = 1; i < n; i++) {
for (int j = 0; j < 9; j++) {
for (int k = 0; k < 9; k++) {
cin >> arr[j][k];
}
}
brr[i] = identify();
a = getchar();
}
for (int j = 0; j < 9; j++) {
for (int k = 0; k < 9; k++) {
cin >> arr[j][k];
}
}
brr[n] = identify();
for (int i = 1; i < n; i++) {
if (brr[i] == 1) {
cout << "Right" << endl;
} else {
cout << "Wrong" << endl;
}
}
if (brr[n] == 1) {
cout << "Right" ;
} else {
cout << "Wrong" ;
}
return 0;
}

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;

namespace dts
{
    int n,a[9][9],vis[10];
    
    int check()
    {
        for (int i=0;i<9;i++)
        {
            memset(vis,0,sizeof(vis));
            for (int j=0;j<9;j++)
                if (1<=a[i][j]&&a[i][j]<=9&&vis[a[i][j]]==0)
                    vis[a[i][j]]=1;
                else
                    return 0;
        }
        for (int j=0;j<9;j++)
        {
            memset(vis,0,sizeof(vis));
            for (int i=0;i<9;i++)
                if (1<=a[i][j]&&a[i][j]<=9&&vis[a[i][j]]==0)
                    vis[a[i][j]]=1;
                else
                    return 0;
        }
        for (int i=0;i<9;i+=3)
            for (int j=0;j<9;j+=3)
            {
                memset(vis,0,sizeof(vis));
                for (int k=i;k<=i+2;k++)
                    for (int l=j;l<=j+2;l++)
                        if (1<=a[k][l]&&a[k][l]<=9&&vis[a[k][l]]==0)
                            vis[a[k][l]]=1;
                        else
                            return 0;
            }
        return 1;
    }
    
    void main()
    {
        scanf("%d",&n);
        for (int rp=1;rp<=n;rp++)
        {
            for (int i=0;i<9;i++)
                for (int j=0;j<9;j++)
                    scanf("%d",&a[i][j]);
            if (check())
                printf("Right\n");
            else
                printf("Wrong\n");
        }
    }
}

int main()
{
    dts::main();
}

算法竞赛入门经典里有句话叫 Keep It Simple and Stupid(KISS)，我觉得特别适合这道题。
考场上最容易想到且最好实现的算法才是最靠谱的。这题思路就是每次把9个数加到vector里，然后用标准库里的find函数看是不是1-9都出现了一遍，这样可以少写很多代码而且这个思路写出来也不容易出bug。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int arr[9][9];
vector<int> v;

// 检查vector里是不是1-9九个数都有
bool check_vector() {
    for (int i = 1; i <= 9; ++i) {
        if (find(v.begin(), v.end(), i) == v.end()) {
            return 0;
        }
    }
    return 1;
}

bool check() {
    // 检查行
    for (int i = 0; i < 9; ++i) {
        v.clear();
        for (int j = 0; j < 9; ++j) {
            v.push_back(arr[i][j]);
        }
        if (!check_vector())
            return 0;
    }
    // 检查列
    for (int i = 0; i < 9; ++i) {
        v.clear();
        for (int j = 0; j < 9; ++j) {
            v.push_back(arr[j][i]);
        }
        if (!check_vector())
            return 0;
    }
    // 检查方格
    for (int i = 0; i < 3; ++i) {
        for (int j = 0; j < 3; ++j) {
            v.clear();
            for (int k = 0; k < 3; ++k) {
                for (int p = 0; p < 3; ++p) {
                    v.push_back(arr[3 * i + k][3 * j + p]);
                }
            }
            if (!check_vector())
                return 0;
        }
    }
    return 1;
}

int main(int argc, char *argv[]) {
#ifdef DEBUG
    freopen("a.in", "r", stdin);
#endif
    int n;
    cin >> n;

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < 9; ++j) {
            for (int k = 0; k < 9; ++k) {
                cin >> arr[j][k];
            }
        }
        if (check())
            cout << "Right" << endl;
        else
            cout << "Wrong" << endl;
    }

    return 0;
}

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<ctime>
#include<algorithm>
#include<iomanip>
#define f(i, a, b) for(int i = a; i <=  b; i++)
using namespace std;
const int N = 15;
int a[N][N];
int l[N][N] = 
{
    {0,0,0,0,0,0,0,0,0,0},
    {0,1,1,1,2,2,2,3,3,3},
    {0,1,1,1,2,2,2,3,3,3},
    {0,1,1,1,2,2,2,3,3,3},
    {0,4,4,4,5,5,5,6,6,6},
    {0,4,4,4,5,5,5,6,6,6},
    {0,4,4,4,5,5,5,6,6,6},
    {0,7,7,7,8,8,8,9,9,9},
    {0,7,7,7,8,8,8,9,9,9},
    {0,7,7,7,8,8,8,9,9,9}
};
bool m1[N][N], m2[N][N], m3[N][N];
int t;
int main()
{
    int m;bool sd;
    scanf("%d", &t);
    while(t--){
        sd = true;
        f(i, 1, 9){
            f(j, 1, 9){
                scanf("%d", &m);
                if(m1[i][m] || m2[j][m] || m3[l[i][j]][m]){
                    sd = false;
                    break;
                }
                m1[i][m] = m2[j][m] = m3[l[i][j]][m] = true;
            }
            if(!sd)break;
        }
        if(!sd)printf("Wrong\n");
        else printf("Right\n");
    }
    
    return 0;
}
**自己拿去调试**

简单模拟

#include <iostream>
using namespace std;

int i, j, k;
int yans[10];//模拟标记，记录1~9的数字并验证
                         //可以多开几个标记数组共同验证，达到减少复杂度的目的

bool toyan() {
  int k = 1, flag = 1;
  for (k = 1; k < 10; k++) {
    if (yans[k] != 1)
      flag = 0;
    yans[k] = 0; //清除标记
  }
  if (flag)
    return true;
  else
    return false;
}

bool yanzheng(int a[9][9]) {
  for (i = 0; i < 9; i++) {
    //验证列
    for (j = 0; j < 9; j++) {
      yans[a[j][i]] = 1;
    }
    if (!toyan())
      return false;
    //验证行
    for (j = 0; j < 9; j++) {
      yans[a[i][j]] = 1;
    }
    if (!toyan())
      return false;
  }

  //验证3*3方块
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 3; j++) {
      for (int m = 3 * i; m < 3 * i + 3; m++) {
        for (int n = 3 * j; n < 3 * j + 3; n++) {
          yans[a[m][n]] = 1;
        }
      }
      if (!toyan())
        return false;
    }
  }
  return true;
}

int main() {
  int a[9][9];
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (j = 0; j < 81; j++) {
      cin >> a[j / 9][j % 9];
    }
    getchar();
    if (yanzheng(a))
      cout << "Right" << endl;
    else
      cout << "Wrong" << endl;
  }
  getchar();
}

#include<iostream>
using namespace std;
int a[100][10][10];
int main()
{
int n,x,y,z=0;
cin>>n;
for(int i=0;i<n;i++)
for(int j=0;j<9;j++)
for(int k=0;k<9;k++)
cin>>a[i][j][k];
for(int i=0;i<n;i++)
for(int j=0;j<9;j++)
{
for(int k=0;k<9;k++)
{
for(int m=k+1;m<9;m++)
if((a[i][j][k]==a[i][j][m])||(a[i][j][k]==a[i][m][k]))
{
z=1;
break;
}
int e=(j/3)*3+1;
int f=(k/3)*3+1;
for(int c=e-1;c<=e+1;c++)
{
for(int d=f-1;d<=f+1;d++)
if((a[i][j][k]==a[i][c][d])&&((j!=c)&&(k!=d)))
{
z=1;
break;
}
if(z==1) break;
}
if(z==1) break;
}
if(z==1)
{
z=0;
cout<<"Wrong"<<endl;
break;
}
else
{
z=0;
cout<<"Right"<<endl;
break;
}
}
return 0;
}

#include<bits/stdc++.h>
using namespace std;
bool check(int a[10][10])
{
int b[10]={0};
int cc=4;
for(int x=0;x<9;x++)
{
for(int y=0;y<9;y++)
{
switch(a[x][y])
{
case 1:b[1]+=1;break;
case 2:b[2]+=1;break;
case 3:b[3]+=1;break;
case 4:b[4]+=1;break;
case 5:b[5]+=1;break;
case 6:b[6]+=1;break;
case 7:b[7]+=1;break;
case 8:b[8]+=1;break;
case 9:b[9]+=1;break;

}
}
for(int t=1;t<=9;t++)
if(b[t]!=x+1)
cc= 0;
}
memset(b,0,sizeof(b));
for(int y=0;y<9;y++)
{
for(int x=0;x<9;x++)
{
switch(a[x][y])
{
case 1:b[1]+=1;break;
case 2:b[2]+=1;break;
case 3:b[3]+=1;break;
case 4:b[4]+=1;break;
case 5:b[5]+=1;break;
case 6:b[6]+=1;break;
case 7:b[7]+=1;break;
case 8:b[8]+=1;break;
case 9:b[9]+=1;break;
}
}
for(int t=1;t<=9;t++)
if(b[t]!=y+1)
cc= 0;
}
int sum=0;
for(int x=0;x<9;x++)
{
for(int y=0;y<9;y++)
sum = sum+a[x][y];
if(sum!=45)
cc= 0;
sum =0;
}
for(int y=0;y<9;y++)
{
for(int x=0;x<9;x++)
sum = sum+a[x][y];
if(sum!=45)
cc= 0;
sum=0;
}
sum=0;
for(int x=0;x<3;x++)
{
for(int y=0;y<3;y++)
sum=sum+a[x][y];
}
if(sum!=45)
cc=0;
if(cc!=0)
cc=1;
return cc;
}
int main()
{ int n;
int a[10][10];
cin>>n;
for(int t=0;t<n;t++)
{

for(int i=0;i<9;i++)
{
for(int j=0;j<9;j++)
{
cin>>a[i][j];
}
}
if(check(a)==0)
cout<<"Wrong"<<endl;
else
cout<<"Right"<<endl;
memset(a,0,sizeof(a));
}
return 0;
}

#include<bits/stdc++.h>
void judge(int n,int a[][10][10]){
    bool flag[10]={0};
    for(int i=1;i<=9;++i){
        for(int j=1;j<=9;++j) if(1^flag[a[n][i][j]]) flag[a[n][i][j]]=1;else{printf("Wrong\n");return;}
        memset(flag,0,sizeof(flag));
    }
    for(int i=1;i<=9;++i){
        for(int j=1;j<=9;++j) if(1^flag[a[n][i][j]]) flag[a[n][i][j]]=1;else{printf("Wrong\n");return;}
        memset(flag,0,sizeof(flag));
    }
    for(int i=0;i<=6;i=i+3)
        for(int j=0;j<=6;j=j+3){
            for(int k=1;k<=3;++k)
                for(int l=1;l<=3;++l)
                    if(1^flag[a[n][i+k][j+l]]) flag[a[n][i+k][j+l]]=1;else{printf("Wrong\n");return;}
            memset(flag,0,sizeof(flag));
        }
    printf("Right\n");
}
int main(){
    int n,a[21][10][10];
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
        for(int j=1;j<=9;++j)
            for(int k=1;k<=9;++k)
                scanf("%d",&a[i][j][k]);
    for(int i=1;i<=n;++i) judge(i,a);
}

/*Tokgo*/
/*
没什么技巧
纯模拟
当然可以用一维数组
舒服一点
*/
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
int m[82];
bool z[9][10],h[9][10];
bool k[3][3][10];
int zj[9],hj[9],kj[3][3];
int main(){
    int n;
    cin>>n;
    while(n--){
        bool jud=1;
        memset(z,0,sizeof(z));
        memset(h,0,sizeof(h));
        memset(k,0,sizeof(k));
        memset(zj,0,sizeof(k));
        memset(hj,0,sizeof(k));
        memset(kj,0,sizeof(k));
        for(int i=1;i<=81;++i){
            cin>>m[i];      
            if(!z[i%9][m[i]]){
                z[i%9][m[i]]=1;
                zj[i%9]++;
            }
            else jud=0;
            if(!h[(i-1)/9][m[i]]){
                h[(i-1)/9][m[i]]=1;
                ++hj[(i-1)/9];
            }
            else jud=0;
            if(!k[(((i-1)%9))/3][((i-1)/9)/3][m[i]]){
                k[(((i-1)%9))/3][((i-1)/9)/3][m[i]]=1;
                ++kj[(((i-1)%9))/3][((i-1)/9)/3];
            }
            else jud=0;
        }
        for(int i=0;i<9;++i){
            if(zj[i]!=9||hj[i]!=9)
                jud=0;
        }
        for(int i=0;i<3;++i){
            for(int j=0;j<3;++j){
                if(kj[i][j]!=9)
                    jud=0;
            }
        }
        if(jud)
            cout<<"Right"<<endl;
        else
            cout<<"Wrong"<<endl;
    }
    return 0;
}

python3代码，略显简陋但按部就班!:-)

def judge(s):
    flag = False
    for i in range(0,9,3):
        a,b,c = [],[],[]
        for j in range(i,i+3):
            a.extend(s[j][0:3])
            b.extend(s[j][3:6])
            c.extend(s[j][6:9])
        if len(set(a))+len(set(b))+len(set(c)) != 27:
            flag = True
            break
    return flag
        
n = int(input())
for i in range (n):
    s = [];flag = False
    s.append(list(map(int,input().split())))
    if not s[0]: # 空行
        s[0] = (list(map(int,input().split())))
    for j in range(8):
        s.append(list(map(int,input().split())))
    s2 = list(map(list, zip(*s))) # 利用解包 矩阵转置
    for j in range(9): # 行列检验
        if len(set(s[i]))+len(set(s2[i])) != 18:
            flag = True
            break
    #九宫格检验
    flag = flag or judge(s)
    print(["Right","Wrong"][flag])
    

我觉得我的最简单暴力，但是代码比较多，复杂度比较高些

#include<bits/stdc++.h>
using namespace std;
int maps[10][10];
bool used[10];
bool chackused() {
    for (int i = 1; i < 10; i++)
    {
        if (!used[i]) return false;
    }
    return true;
}
bool chackrow() {
    for (int i = 1; i < 10 ; i++)
    {
        memset(used, false, sizeof(used));
        for (int j = 1; j < 10; j++)
        {
            used[maps[i][j]] = true;
        }
        if (!chackused()) return false;
    }
    return true;
}
bool chackcol() {
    for (int j = 1; j < 10; j++){
        memset(used, false, sizeof(used));
        for (int i = 1; i < 10; i++)
        {
            used[maps[i][j]] = true;
        }
        if (!chackused()) return false;
    }
    return true;
}
bool chack33(int ai,int bj) {
    memset(used, false, sizeof(used));
    for (int i = ai; i < ai+3 ; i++)
    {
        for (int j = bj; j < bj+3; j++)
        {
            used[maps[i][j]] = true;
        }
    }
    return chackused();
}
bool chack99() {
    for (int i = 1; i < 10; i+=3)
    {
        for (int j = 1; j < 10 ; j+=3)
        {
            if (!chack33(i, j)) return false;
        }
    }
    return true;
}
int main() {
    int n;
    cin >> n;
    while (n--)
    {
        for (int i = 1; i < 10; i++)
        {
            for (int j = 1; j < 10; j++)
            {
                cin >> maps[i][j];
            }
        }
        if (chackrow() && chackcol() && chack99()) cout << "Right" << endl;
        else cout << "Wrong" << endl;
    }
    return 0;
}

无非是测试横竖以及9个方格内是否出现了全部的1~9。那么如何验证是否出现了全部的1~9呢？可以用乘法。9的阶乘是362880，如果相乘的不是这9个数字，那么乘积一定不是362880。
```cpp
#include <iostream>
using namespace std;

int main()
{
int n;
cin >> n;
int a[9][9];
bool iswrong;
while (n--)
{
iswrong = false;
for (int i = 0; i < 9; i++)
{
for (int j = 0; j < 9; j++)
cin >> a[i][j];
}
for (int i = 0; i < 9; i++)
{
if (a[i][0] * a[i][1] * a[i][2] * a[i][3] * a[i][4] * a[i][5] * a[i][6] * a[i][7] * a[i][8] != 362880)
{
cout << "Wrong" << endl;
iswrong = true;
break;
}
if (a[0][i] * a[1][i] * a[2][i] * a[3][i] * a[4][i] * a[5][i] * a[6][i] * a[7][i] * a[8][i] != 362880)
{
cout << "Wrong" << endl;
iswrong = true;
break;
}
}
if (!iswrong)
{
int multi;
for (int i = 0; i < 9; i += 3)
{
multi = 1;
for (int j = 0; j < 3; j++)
{
for (int k = 0; k < 3; k++)
multi *= a[i + j][i + k];
}
if (multi != 362880)
{
cout << "Wrong" << endl;
iswrong = true;
break;
}
}
}
if (!iswrong)
cout << "Right" << endl;
}
return 0;
}
```

//数独验证
#include<iostream>
using namespace std;
#include<algorithm>
#include<stdlib.h>
#include<string.h>

int matrix[15][15] = { 0 };
int n;
bool row_check() {
int flag[15] = { 0 };
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (flag[matrix[i][j]] >= 2)
return false;
flag[matrix[i][j]]++;
}
memset(flag, 0, sizeof(flag));
}
return true;
}

bool column_check() {
int flag[15] = { 0 };
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (flag[matrix[j][i]] >= 2)
return false;
flag[matrix[j][i]]++;
}
memset(flag, 0, sizeof(flag));
}
return true;
}

bool block_check() {
int flag[15] = { 0 };
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
for (int k = 0; k < 3; k++) {
for (int l = 0; l < 3; l++) {
flag[matrix[i * 3 + k][j * 3 + l]]++;
}
}
for(int k = 1; k <= 9; k++)
if (!flag[k])
return false;
memset(flag, 0, sizeof(flag));
}
}
return true;
}

int main()
{
cin >> n;
while (n--) {
for (int i = 0; i < 9; i++)
for (int j = 0; j < 9; j++)
cin >> matrix[i][j];
getchar();
if (row_check() && column_check() && block_check())
cout << "Right" << endl;
else cout << "Wrong" << endl;
}
return 0;
}

这题有坑 正常人谁数独会填0的