label 1;
var flag,n,i,j,k,time1,time2,sum1,sum2:longint;
a:array[1..10,1..10] of longint;
begin
readln(n);
for i:=1 to n do begin
flag:=1;
for j:=1 to 9 do begin
for k:=1 to 9 do read(a[j,k]);
readln();
end;
if i<>n then readln();

for j:=1 to 9 do begin
time1:=1;
time2:=1;
sum1:=0;
sum2:=0;
for k:=1 to 9 do begin
time1:=time1*a[j,k];
time2:=time2*a[k,j];
sum1:=sum1+a[j,k];
sum2:=sum2+a[k,j];
end;

if (time1<>362880) or (sum1<>45) then goto 1;
if (time2<>362880) or (sum2<>45) then goto 1;
end;
time1:=1;
sum1:=0;

for j:=1 to 3 do begin
for k:=1 to 3 do begin
time1:=time1*a[j,k];
sum1:=sum1+a[j,k];
end;
end;
if (time1<>362880) or (sum1<>45) then goto 1;

time1:=1;
sum1:=0;

for j:=1 to 3 do begin
for k:=4 to 6 do begin
time1:=time1*a[j,k];
sum1:=sum1+a[j,k];
end;
end;
if (time1<>362880) or (sum1<>45) then goto 1;

time1:=1;
sum1:=0;

for j:=1 to 3 do begin
for k:=7 to 9 do begin
time1:=time1*a[j,k];
sum1:=sum1+a[j,k];
end;
end;
if (time1<>362880) or (sum1<>45) then goto 1;

time1:=1;
sum1:=0;

for j:=4 to 6 do begin
for k:=1 to 3 do begin
time1:=time1*a[j,k];
sum1:=sum1+a[j,k];
end;
end;
if (time1<>362880) or (sum1<>45) then goto 1;

time1:=1;
sum1:=0;

for j:=4 to 6 do begin
for k:=4 to 6 do begin
time1:=time1*a[j,k];
sum1:=sum1+a[j,k];
end;
end;
if (time1<>362880) or (sum1<>45) then goto 1;

time1:=1;
sum1:=0;

for j:=4 to 6 do begin
for k:=7 to 9 do begin
time1:=time1*a[j,k];
sum1:=sum1+a[j,k];
end;
end;
if (time1<>362880) or (sum1<>45) then goto 1;

time1:=1;
sum1:=0;

for j:=7 to 9 do begin
for k:=1 to 3 do begin
time1:=time1*a[j,k];
sum1:=sum1+a[j,k];
end;
end;
if (time1<>362880) or (sum1<>45) then goto 1;

time1:=1;
sum1:=0;

for j:=7 to 9 do begin
for k:=4 to 6 do begin
time1:=time1*a[j,k];
sum1:=sum1+a[j,k];
end;
end;
if (time1<>362880) or (sum1<>45) then goto 1;

time1:=1;
sum1:=0;

for j:=7 to 9 do begin
for k:=7 to 9 do begin
time1:=time1*a[j,k];
sum1:=sum1+a[j,k];
end;
end;
if (time1<>362880) or (sum1<>45) then goto 1;

time1:=1;
sum1:=0;

for j:=1 to 3 do begin
for k:=1 to 3 do begin
time1:=time1*a[j,k];
sum1:=sum1+a[j,k];
end;
end;
if (time1<>362880) or (sum1<>45) then goto 1;

flag:=0;
1:if flag=1 then writeln('Wrong') else writeln('Right');
end;
end.

我先求和，但没通过。后来用集合，才通过的。
var
a:array[1..9,1..9]of integer;
i,j,n,k,s,l:integer;
b:boolean;
t:set of 1..9;
begin
read(n);
for k:=1 to n do
begin
b:=true;
for i:=1 to 9 do
for j:=1 to 9 do read(a[i,j]);
for i:=1 to 9 do
begin
t:=[];
for j:=1 to 9 do
if not(a[i,j]in t) then t:=t+[a[i,j]]
else b:=false;
end;
for j:=1 to 9 do
begin
t:=[];
for i:=1 to 9 do
if not(a[i,j]in t) then t:=t+[a[i,j]]
else b:=false;
end;
for i:=1 to 3 do
for j:=1 to 3 do
begin
t:=[];
t:=t+[a[i*3-2,j*3-2]]+[a[i*3-1,j*3-2]]+[a[i*3,j*3-2]]+[a[i*3-2,j*3-1]]+[a[i*3-1,j*3-1]]+[a[i*3,j*3-1]]+[a[i*3-2,j*3]]+[a[i*3-1,j*3]]+[a[i*3,j*3]];
s:=0;
for l:=1 to 9 do
if (l in t) then s:=s+1;
if s<>9 then b:=false;
end;
if b then writeln('Right')
else writeln('Wrong');
end;
end.

30行，超时。。。

program p1335;

var a:array[1..9,1..9] of 1..9;

num:array[1..27] of set of 1..9;

n,i,j,k,p:longint;

flag:array[1..10] of boolean;

procedure check;

begin

for i:=1 to 27 do num[i]:=[1..9];

for i:=1 to 9 do

begin

for j:=1 to 9 do

begin

read(a);

num[i]:=num[i]-[a];

num[9+j]:=num[9+j]-[a];

p:=((i-1) div 3)*3+(j-1) div 3+19;

num[p]:=num[p]-[a];

end;

readln;

end;

for i:=1 to 27 do if num[i][] then

begin

flag[k]:=false;

break;

end;

end;

begin

readln(n);

for k:=1 to n do begin flag[k]:=true;check;end;

for k:=1 to n do if flag[k] then writeln('Right')

else writeln('Wrong');

end.

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

program p1335;

var a,sum:array[0..9,0..9] of longint;

v:array[0..9] of boolean;

i,j,k,l,m,n:longint;

function check:boolean;

var g,t,s,l,r:longint;

begin

for g:=1 to 9 do begin fillchar(v,sizeof(v),false);

for t:=1 to 9 do

if v[a[g,t]]=true then exit(false) else v[a[g,t]]:=true;

end;

for g:=1 to 9 do begin fillchar(v,sizeof(v),false);

for t:=1 to 9 do

if v[a[t,g]]=true then exit(false) else v[a[t,g]]:=true;

end;

for l:=0 to 2 do

for r:=0 to 2 do

begin

fillchar(v,sizeof(v),false);

for g:=1 to 3 do

for t:=1 to 3 do

if v[a[l*3+g,r*3+t]] then exit(false) else v[a[l*3+g,r*3+t]]:=true;

end;

exit(true);

end;

begin

readln(n);

for k:=1 to n do

begin

for i:=1 to 9 do

for j:=1 to 9 do read(a);

if check then writeln('Right') else writeln('Wrong');

end;

end.

光用和的话会导致第四个数据出错，还要判断积是不是等于362880

var a:array[1..9,1..9] of byte;

i,j,k,l,p:byte;

s1,s2,s3:integer;

sl,sh,sg:longint;

n:integer;

function find:boolean;

begin

fillchar(a,sizeof(a),0);

s1:=0;s2:=0;

for i:=1 to 9 do begin

for j:=1 to 9 do

read(a);

readln;

end;

for i:=1 to 9 do begin

s1:=0;s2:=0;sl:=1;sh:=1;

for j:=1 to 9 do

begin

s1:=s1+a;

sl:=sl*a;

sh:=sh*a[j,i];

s2:=s2+a[j,i];

end;

if (s145)or(s245)or(sl362880)or(sh362880) then begin find:=false;

exit;

end;

end;

for l:=1 to 3 do

for p:=1 to 3 do begin s3:=0;sl:=1;

for i:=1+(l-1)*3 to l*3 do

for j:=1+(p-1)*3 to P*3 do

begin s3:=s3+a;sl:=sl*a;end;

if (s345)or(sl362880) then begin find:=false;exit;end;

end;

find:=true;

end;

begin

readln(n);

for k:=1 to n do begin

if find then writeln('Right') else writeln('Wrong');

readln;

end;

end.

1次秒杀...

这种情况现在对我来说貌似很少见了

庆祝一下：

#include

int n;

main()

{

scanf("%d",&n);

for(;n>0;n--)

{

int a[9][9],i=0,j=0,flag=0;

for(;i

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

var a:array [1..9,1..9] of integer;

b,c:array [1..9] of boolean;

i,j,k,p,q,r,s:Integer;

n:integer;

result:boolean;

begin

read(n);

for i:=1 to n do

begin

fillchar(a,sizeof(a),0);

result:=true;

for p:=1 to 9 do

for q:=1 to 9 do

begin

read(a[p,q]);

end;

for p:=1 to 9 do

begin

fillchar(b,sizeof(b),false);

fillchar(c,sizeof(c),false);

for q:=1 to 9 do

begin

if b[a[p,q]]=false then b[a[p,q]]:=true

else begin result:=false;break;end;

if c[a[q,p]]=false then c[a[q,p]]:=true

else begin result:=false;break;end;

end;

if result=false then break;

end;

if result=true then

for p:=0 to 2 do

begin

for q:=0 to 2 do

begin

fillchar(b,sizeof(b),false);

for r:=1 to 3 do

begin

for s:=1 to 3 do

if b[a[p*3+r,q*3+s]]=false then b[a[p*3+r,q*3+s]]:=true

else begin result:=false;break;end;

if result=false then break;

end;

if result=false then break;

end;

if result=false then break;

end;

if result=true then writeln('Right') else writeln('Wrong');

end;

end.

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

很简单 50+ 一次ac

晕，交了两次....记住，要读完再跳出来！

---|---|---|---|---|---|---|---|---|---|---|-

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

const num:array[1..9]of longint=(1,2,4,8,16,32,64,128,256);

all=511;

var n,i,j:longint;

function init:boolean;

var a,i,j,k:longint;

h,z,gong:array[1..9]of longint;

begin

fillchar(z,sizeof(z),0);

fillchar(h,sizeof(h),0);

fillchar(gong,sizeof(gong),0);

for i:=1 to 9 do

begin

for j:=1 to 9 do

begin

read(a);

k:=((i-1) div 3)*3+((j-1) div 3)+1;

gong[k]:=gong[k] or num[a];

h[i]:=h[i] or num[a];

z[j]:=z[j] or num[a];

end;

readln;

end;

for i:=1 to 9 do

begin

if z[i]all then exit(false);

if gong[i]all then exit(false);

if h[i]all then exit(false);

end;

exit(true);

end;

begin

readln(n);

for i:=1 to n do

begin

if init then writeln('Right') else writeln('Wrong');

readln;

end;

end.

好水啊 一次AC

program p1335;

var

n :integer;

a :array[1..9,1..9] of integer;

b :array[1..9] of boolean;

m :integer;

i,j :integer;

flag :boolean;

procedure make(p,q:integer);

var s :integer;

begin

b[a[p,q]]:=false;

b[a[p+1,q]]:=false;

b[a[p+2,q]]:=false;

b[a[p,q+1]]:=false;

b[a[p+1,q+1]]:=false;

b[a[p+2,q+1]]:=false;

b[a[p,q+2]]:=false;

b[a[p+1,q+2]]:=false;

b[a[p+2,q+2]]:=false;

for s:=1 to 9 do

if b=true then flag:=true;

end;

begin

readln(n);

for m:=1 to n do

begin

for i:=1 to 9 do

for j:=1 to 9 do

read(a);

flag:=false;

for i:=1 to 9 do

begin

if flag=true then break;

fillchar(b,sizeof(b),true);

for j:=1 to 9 do

begin

b[a]:=false;

end;

for j:=1 to 9 do

if b[j]=true then begin flag:=true;break;end;

end;

for i:=1 to 3 do

begin

if flag=true then break;

for j:=1 to 3 do

begin if flag=true then break;

fillchar(b,sizeof(b),true);

make(3*i-2,3*j-2);

end;

end;

if flag=true then writeln('Wrong') else writeln('Right');

end;

end.

"没有读完一个数独的数而又已经能判断其为‘WRONG’时，直接跳到第2个数独，可能产生第2个和第9个测试数据错误"

WA了3次……

他喵的.....我恨

我交第一遍的时候过4个点，然后我用数据自己测，觉得输出没错， 仔细一看

原来 我把 right 打成了 rignt

不容易看出来吧.......

我晕，居然就没看到前面还有一个n。

偶的通过率啊！

这题怎么就会多个n呢！！！！！！

不过这样也能过个点，怪吧…………

看到我的最短，发上来让大家看看思路吧

var i,j,k,n:longint;

cols,rows,boxs:array[0..10,0..10] of boolean; bool:boolean;

procedure check;

var i,j,d:longint;

begin

bool:=false;

fillchar(cols,sizeof(cols),false);

fillchar(rows,sizeof(rows),false);

fillchar(boxs,sizeof(boxs),false);

for i:=1 to 9 do

begin

for j:=1 to 9 do

begin

read(d);

if (not(cols[j,d])) and (not(rows)) and (not(boxs[(((i-1) div 3)*3)+((j+2) div 3),d])) and (d0) then

begin

cols[j,d]:=true;

rows:=true;

boxs[(((i-1) div 3)*3)+((j+2) div 3),d]:=true;

end

else bool:=true;

end;

end;

if bool then writeln('Wrong') else writeln('Right');

end;

begin

readln(n);

for i:=1 to n do

begin

check;

end;

end.

var

a,b,c:array[1..9,1..9]of boolean;

n,i:longint;

procedure init;

begin

readln(n);

end;

procedure main;

var

i,j,k,tmp:longint;

flag:boolean;

begin

fillchar(a,sizeof(a),false);

fillchar(b,sizeof(b),false);

fillchar(c,sizeof(c),false);

tmp:=0;

k:=0;

for i:=1 to 9 do

begin

for j:=1 to 9 do

begin

read(tmp);

if (i>=1) and (i=1) and (j=4) and (j=7) and (j=4) and (i=1) and (j=4) and (j=7) and (j=7) and (i=1) and (j=4) and (j=7) and (j

巨猥琐的题目 336行！整整10KB

话说用集合做比较不错

type

dare=set of 1..9;

var

a:array[1..3,1..3] of dare;

b,c:array[1..9] of dare;

i,j,n,p,q,k,w,h:integer;

boo1,boo2:boolean;

begin

readln(n);

for h:=1 to n do

begin

for i:=1 to 3 do

for j:=1 to 3 do

a:=[];

for i:=1 to 9 do

begin

b[i]:=[];

c[i]:=[];

end;

for j:=1 to 9 do

begin

for k:=1 to 9 do

begin

read(w);

b[j]:=b[j]+[w];

c[k]:=c[k]+[w];

if (j/3)

我就140行哈~~

不断复制程序就可以了...

就是简单的模拟啦

228行，巨繁无比。

program p1335;

var n,i,j,k,l,s:integer;

t:boolean;

a:array[1..9,1..9] of 1..9;

b:array[1..9] of 1..9;

begin

read(n);

for i:=1 to n do

begin

for j:=1 to 9 do

for k:=1 to 9 do read(a[j,k]);

t:=true;

for j:=1 to 9 do

begin

for k:=1 to 8 do

begin

for l:=k+1 to 9 do

if a[j,k]=a[j,l] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then break;

end;

if not t then continue;

t:=true;

for k:=1 to 9 do

begin

for j:=1 to 8 do

begin

for l:=j+1 to 9 do

if a[j,k]=a[l,k] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then break;

end;

if not t then continue;

s:=0;

for j:=1 to 3 do

for k:=1 to 3 do

begin

s:=s+1;

b:=a[j,k];

end;

t:=true;

for j:=1 to 8 do

begin

for k:=j+1 to 9 do

if b[j]=b[k] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then continue;

s:=0;

for j:=1 to 3 do

for k:=4 to 6 do

begin

s:=s+1;

b:=a[j,k];

end;

t:=true;

for j:=1 to 8 do

begin

for k:=j+1 to 9 do

if b[j]=b[k] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then continue;

s:=0;

for j:=1 to 3 do

for k:=7 to 9 do

begin

s:=s+1;

b:=a[j,k];

end;

t:=true;

for j:=1 to 8 do

begin

for k:=j+1 to 9 do

if b[j]=b[k] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then continue;

s:=0;

for j:=4 to 6 do

for k:=1 to 3 do

begin

s:=s+1;

b:=a[j,k];

end;

t:=true;

for j:=1 to 8 do

begin

for k:=j+1 to 9 do

if b[j]=b[k] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then continue;

s:=0;

for j:=4 to 6 do

for k:=4 to 6 do

begin

s:=s+1;

b:=a[j,k];

end;

t:=true;

for j:=1 to 8 do

begin

for k:=j+1 to 9 do

if b[j]=b[k] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then continue;

s:=0;

for j:=4 to 6 do

for k:=7 to 9 do

begin

s:=s+1;

b:=a[j,k];

end;

t:=true;

for j:=1 to 8 do

begin

for k:=j+1 to 9 do

if b[j]=b[k] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then continue;

s:=0;

for j:=7 to 9 do

for k:=1 to 3 do

begin

s:=s+1;

b:=a[j,k];

end;

t:=true;

for j:=1 to 8 do

begin

for k:=j+1 to 9 do

if b[j]=b[k] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then continue;

s:=0;

for j:=7 to 9 do

for k:=4 to 6 do

begin

s:=s+1;

b:=a[j,k];

end;

t:=true;

for j:=1 to 8 do

begin

for k:=j+1 to 9 do

if b[j]=b[k] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then continue;

s:=0;

for j:=7 to 9 do

for k:=7 to 9 do

begin

s:=s+1;

b:=a[j,k];

end;

t:=true;

for j:=1 to 8 do

begin

for k:=j+1 to 9 do

if b[j]=b[k] then

begin

t:=false;

writeln('Wrong');

break;

end;

if not t then break;

end;

if not t then continue;

writeln('Right');

end;

end.

program das;

var n,i1,i,j,x,y:integer;

a:array[1..9,1..9] of integer;

b:array[1..9] of integer;

f1,f2,f3:boolean;

begin

assign(input,'p1335.in');assign(output,'p1335.out');

reset(input);rewrite(output);

read(n);

for i1:=1 to n do

begin

for i:=1 to 9 do

for j:=1 to 9 do read(a);

for i:=1 to 9 do

begin

fillchar(b,sizeof(b),0);

f1:=true;

for j:=1 to 9 do if b[a]=0 then b[a]:=1

else

begin

writeln('Wrong');

f1:=false;

break;

end;

if not(f1) then break;

end;

if f1 then

begin

for i:=1 to 9 do

begin

fillchar(b,sizeof(b),0);

f2:=true;

for j:=1 to 9 do if b[a[j,i]]=0 then b[a[j,i]]:=1

else

begin

writeln('Wrong');

f2:=false;

break;

end;

if not(f2) then break;

end;

end;

if f1 and f2 then

begin

fillchar(b,sizeof(b),0);

f3:=true;

for i:=1 to 3 do

begin

for j:=1 to 3 do if b[a]=0 then b[a]:=1

else

begin

writeln('Wrong');

f3:=false;

break;

end;

if not(f3) then break;

end;

if f3 then

begin

fillchar(b,sizeof(b),0);

f3:=true;

for i:=1 to 3 do

begin

for j:=4 to 6 do if b[a]=0 then b[a]:=1

else

begin

writeln('Wrong');

f3:=false;

break;

end;

if not(f3) then break;

end;

end;

if f3 then

begin

fillchar(b,sizeof(b),0);

f3:=true;

for i:=1 to 3 do

begin

for j:=7 to 9 do if b[a]=0 then b[a]:=1

else

begin

writeln('Wrong');

f3:=false;

break;

end;

if not(f3) then break;

end;

end;

if f3 then

begin

fillchar(b,sizeof(b),0);

f3:=true;

for i:=4 to 6 do

begin

for j:=1 to 3 do if b[a]=0 then b[a]:=1

else

begin

writeln('Wrong');

f3:=false;

break;

end;

if not(f3) then break;

end;

end;

if f3 then

begin

fillchar(b,sizeof(b),0);

f3:=true;

for i:=4 to 6 do

begin

for j:=4 to 6 do if b[a]=0 then b[a]:=1

else

begin

writeln('Wrong');

f3:=false;

break;

end;

if not(f3) then break;

end;

end;

if f3 then

begin

fillchar(b,sizeof(b),0);

f3:=true;

for i:=4 to 6 do

begin

for j:=7 to 9 do if b[a]=0 then b[a]:=1

else

begin

writeln('Wrong');

f3:=false;

break;

end;

if not(f3) then break;

end;

end;

if f3 then

begin

fillchar(b,sizeof(b),0);

f3:=true;

for i:=7 to 9 do

begin

for j:=1 to 3 do if b[a]=0 then b[a]:=1

else

begin

writeln('Wrong');

f3:=false;

break;

end;

if not(f3) then break;

end;

end;

if f3 then

begin

fillchar(b,sizeof(b),0);

f3:=true;

for i:=7 to 9 do

begin

for j:=4 to 6 do if b[a]=0 then b[a]:=1

else

begin

writeln('Wrong');

f3:=false;

break;

end;

if not(f3) then break;

end;

end;

if f3 then

begin

fillchar(b,sizeof(b),0);

f3:=true;

for i:=7 to 9 do

begin

for j:=7 to 9 do if b[a]=0 then b[a]:=1

else

begin

writeln('Wrong');

f3:=false;

break;

end;

if not(f3) then break;

end;

end;

end;

if f1 and f2 and f3 then writeln('Right');

readln;

end;

close(input);close(output);

end.

农夫山泉

大把的时间花在敲代码上