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4 x5 w+ c& B; ~4 i0 F* w& t
(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;
}

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2^#

johnson發表於 2011-12-31 11:35 | 只看該作者

public class j102
{
public static void main(String arg[])2 C: V: R1 D6 |( Z' x" ?
{1 a7 I& u" i% A" o
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
{
int r=(int)(Math.random()*9)+1;
int t=num[i];; b% E; ^+ g2 |6 _4 z; Q
num[i]=num[r];5 B/ `- K% r: e/ i
num[r]=t;
int a1 = num[0] + num[1] + num[2] ;9 _; v; O; W! |* ]4 `: C
      int a2 = num[3] + num[4] + num[5] ;1 y3 o, @4 \1 e4 q
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;& Z9 B3 f+ [* C" \/ R
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;- G! W- Y7 i9 \! `+ E6 Q
      int a8 = num[2] + num[4] + num[6] ;3 W5 O+ N- B' d! _
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
      }break;
}
}

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新手上路

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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;
}

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4^#

eric5802發表於 2011-12-31 11:38 | 只看該作者

public class j102
{
public static void main(String arg[])$ `9 h& T- }( ^+ C
      {' A5 Y  T# }$ q- p
         int num []={1,2,3,4,5,6,7,8,9};1 e/ _* R0 t- x
      while(trun)
            {( p" a& F: K4 I/ o
            int r=(int)(Math.random()*9)+1;/ C+ p2 V& h. t: H! q6 [7 e6 j
            int t=num[i];
            num[i]=num[r];
            num[r]=t;) Q7 M3 P+ H7 w; |$ F& F0 L
            int a1 = num[0] + num[1] + num[2] ;5 v+ U- Z- b/ V" F7 L
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;% S& m; Z8 R6 h  ^
      int a4 = num[0] + num[3] + num[6] ;; s7 \5 i2 x7 t& x. m+ m* L- d
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;0 n1 m+ X6 ~7 p6 M
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
         }break;
}
}

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