[C]38th ITSA Problem 2

第二題也還蠻簡單的 其實沒什麼需要注意的地方

#include <stdio.h>

int ooxx[3][3];
void initarray(){
  int i,j;
    for(i=0;i<3;i++)
 for(j=0;j<3;j++)
   ooxx[i][j]=0;
}
void printResult(char winner){
  int i,j;
  //印出井字內容 
  for(i=0;i<3;i++){
    for(j=0;j<3;j++)
 if(ooxx[i][j]==0)
   printf("*");
 else
   printf("%c",ooxx[i][j]==1?'Y':'X');
 printf("\n");
    }
  //印出贏家 
  printf(winner=='T'?"Tie\n":"%c bingle\n",winner);
}
int main(int argc, char *argv[]) {
  int c=0,i;
  char winner[1]={'T'};
  initarray();
  while(scanf("%d %d %d",&ooxx[c][0],&ooxx[c][1],&ooxx[c][2])==3){
    ++c;
    if(c==3){
 //判斷行&列是否bingle & 贏家 
 for(i=0;i<3;i++){
   if(ooxx[i][0]==ooxx[i][1]&&ooxx[i][0]==ooxx[i][2]){
     winner[0]=ooxx[i][0]==1?'Y':'X';
   }else if(ooxx[0][i]==ooxx[1][i]&&ooxx[0][i]==ooxx[2][i]){
     winner[0]=ooxx[0][i]==1?'Y':'X';
   }
 }
 //判斷兩對角線是否bingle & 贏家 
 if(ooxx[0][0]==ooxx[1][1]&&ooxx[0][0]==ooxx[2][2]||ooxx[0][2]==ooxx[1][1]&&ooxx[0][2]==ooxx[2][0])
   winner[0]=ooxx[1][1]==1?'Y':'X';
   //印出 & 初始化陣列&贏家 
   printResult(winner[0]);
   initarray();
   c=0;
   winner[0]='T';
  }
  return 0;
}

以上為個人淺解 還請各位大大多多指教