jva 104 河洛之數

1) 矩陣內數字皆不可重複出現
(2) 程式每次執行，矩陣數字排列皆不相同$ i$ [, G# Y6 G, b! A
(3) 矩陣之縱向、橫向、對角線數字總和皆為15

#include <iostream>
#include <cstdlib>
using namespace std ;
int main(){
int a[] = {1,2,3,4,5,6,7,8,9} ;
//亂數種子
srand(time(NULL));
while(true){
//交換9次
for(int i=0;i<9;i++){
int r = rand()%9 ;
//交換
int t = a[r] ;
a[r] = a[i];
a[i] = t ;
}
//列的判斷
int a1 = a[0] + a[1] + a[2] ;
int a2 = a[3] + a[4] + a[5] ;
int a3 = a[6] + a[7] + a[8] ;
//行的判斷
int a4 = a[0] + a[3] + a[6] ;
int a5 = a[1] + a[4] + a[7] ;
int a6 = a[2] + a[5] + a[8] ;
//斜的判斷
int a7 = a[0] + a[4] + a[8] ;
int a8 = a[2] + a[4] + a[6] ;
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15){
break;
}
}
cout << a[0] << " " << a[1] << " " << a[2] << endl;
cout << a[3] << " " << a[4] << " " << a[5] << endl;
cout << a[6] << " " << a[7] << " " << a[8] << endl;
system("pause");
return 0;
}

複製代碼

新手上路

Rank: 1

2^#

johnson發表於 2011-12-31 11:35 | 只看該作者

public class j102
{* D, V  }2 p" _/ n2 i5 ]- }
public static void main(String arg[])
{
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
{
int r=(int)(Math.random()*9)+1;8 m! ^5 {' c2 f1 j- a' z5 `2 z
int t=num[i];
num[i]=num[r];8 h8 l1 @: K( Q5 ~$ t; r9 Q3 E
num[r]=t;
int a1 = num[0] + num[1] + num[2] ;) C5 K0 M$ H( o& H% Q' }- N. |
      int a2 = num[3] + num[4] + num[5] ;# c5 f( I, a" Z3 X
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;# r& I, O+ G7 h* x
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;  ?5 u& s' E# i: m. o; y1 X" J) A: E
      int a7 = num[0] + num[4] + num[8] ;6 X: Z; V/ Q2 W% l: H  T  J! J
      int a8 = num[2] + num[4] + num[6] ;
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15). u( \+ H; E4 L4 n2 T# n. a" N1 B" O3 Q
      }break;
}
}

小雲雀

TOP

新手上路

Rank: 1

3^#

kim發表於 2011-12-31 11:36 | 只看該作者

#include <iostream>
#include <cstdlib>
using namespace std ;
int main(){
int a[] = {1,2,3,4,5,6,7,8,9} ;
//亂數種子
srand(time(NULL));
while(true){
//交換9次
for(int i=0;i<9;i++){
int r = rand()%9 ;
//交換
int t = a[r] ;
a[r] = a[i];
a[i] = t ;
}
//列的判斷
int a1 = a[0] + a[1] + a[2] ;
int a2 = a[3] + a[4] + a[5] ;
int a3 = a[6] + a[7] + a[8] ;
//行的判斷
int a4 = a[0] + a[3] + a[6] ;
int a5 = a[1] + a[4] + a[7] ;
int a6 = a[2] + a[5] + a[8] ;
//斜的判斷
int a7 = a[0] + a[4] + a[8] ;
int a8 = a[2] + a[4] + a[6] ;
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15){
break;
}
}
cout << a[0] << " " << a[1] << " " << a[2] << endl;
cout << a[3] << " " << a[4] << " " << a[5] << endl;
cout << a[6] << " " << a[7] << " " << a[8] << endl;
system("pause");
return 0;
}

複製代碼

★ 嘉凱~~☆

TOP

中級會員

Rank: 3 Rank: 3

4^#

eric5802發表於 2011-12-31 11:38 | 只看該作者

public class j102- P$ D# ?& f) s; s0 d8 ^$ j
{- @" E5 a0 `0 z% F
public static void main(String arg[])& Q6 H5 U1 p& l! u
      {8 y5 H6 k* c5 z' i4 U& A
         int num []={1,2,3,4,5,6,7,8,9};6 l) F! K+ k  r* M
      while(trun)
            {
            int r=(int)(Math.random()*9)+1;" c& G$ b& Q  f# {7 d) f
            int t=num[i];5 V( ]6 `2 s4 L1 v6 V+ `* c
            num[i]=num[r];* m' ]+ M$ a& B/ i( l
            num[r]=t;0 N. v" Y/ P) n7 g' {- u* d; \( M
            int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;2 y! N$ ?, r+ w5 Y, e2 A
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;2 i0 v) z( E( Y! |  O
      int a6 = num[2] + num[5] + num[8] ;' ^$ z6 u& W. [7 k
      int a7 = num[0] + num[4] + num[8] ;5 R4 O6 R6 I* W. q& m0 t  X( M/ J
      int a8 = num[2] + num[4] + num[6] ;& G$ B3 Y1 H) m7 H4 x0 o( l) r4 q
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)8 i7 _1 Z& |1 I2 `% ]1 C( Y
         }break;
}
}

人平

TOP