Guest User

Untitled

a guest
Jul 20th, 2018
86
0
Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
text 3.63 KB | None | 0 0
  1. #include<stdio.h>
  2. #include<stdlib.h>
  3. #include<string.h>
  4. int main()
  5. {
  6. int i,k,layer;
  7. char ch[82]={'\0'},type='\0',ch2[82]={'\0'};
  8. char way[87385][30]={'\0'},copyway[30]={'\0'};/*way=city*/
  9. /*copyway=ch[2]~ch[16]*/
  10. double jws[87385][3]={'\0'},copy[3]={'\0'};/*0=經度 1=緯度 2=面積*/
  11. int human[87385]={'\0'},copy2=0,treelayer[87385]={'\0'};/*human=人口*/
  12. /*TREELAYER存每個城市的層數*/
  13. while(gets(ch)!=NULL)
  14. {
  15. for(i=0;i<82;i++) /*清array*/
  16. ch2[i]='\0';
  17. for(i=0;i<30;i++)
  18. copyway[i]='\0';
  19. for(i=0;i<3;i++)
  20. copy[i]='\0';
  21. copy2=0;
  22.  
  23. type=ch[0];
  24.  
  25. if(type=='I')
  26. {
  27. for(i=18;i<82;i++) /*CH2=CH的後面那些數字*/
  28. {
  29. ch2[i-18]=ch[i];
  30. }
  31. for(i=2;i<17;i++) /*讓COPYWAY=城市名稱 */
  32. {
  33. copyway[i-2]=ch[i];
  34. }
  35. copy[0]=atof(strtok(ch2," \n\r\t"));/*經度*/
  36. copy[1]=atof(strtok(NULL," \n\r\t"));/*緯度*/
  37. copy2 = atoi(strtok(NULL," \n\r\t"));/*人口*/
  38. copy[2]=atof(strtok(NULL," \n\r\t"));/*面積*/
  39. k=1;
  40. layer=1;/*初始化K 和 LAYER*/
  41. while(treelayer[k]!='\0')
  42. {
  43. if(copy[0]<=jws[k][0] && copy[1]<=jws[k][1])
  44. {
  45. k=4*k-2;
  46. layer++;
  47. }
  48. else if(copy[0]<=jws[k][0] && copy[1]>jws[k][1])
  49. {
  50. k=4*k-1;
  51. layer++;
  52. }
  53. else if(copy[0]>jws[k][0] && copy[1]<=jws[k][1])
  54. {
  55. k=4*k;
  56. layer++;
  57. }
  58. else if(copy[0]>jws[k][0] && copy[1]>jws[k][1])
  59. {
  60. k=4*k+1;
  61. layer++;
  62. }
  63.  
  64. }
  65. /* for(i=0;i<15;i++)
  66. way[k][i]=copyway[i];*/
  67.  
  68.  
  69. strcpy(way[k],copyway); /*把資訊存進ARRAY*/
  70. jws[k][0]=copy[0];
  71. jws[k][1]=copy[1];
  72. human[k]=copy2;
  73. jws[k][2]=copy[2];
  74. treelayer[k]=layer;
  75. printf("OK %s %d\n",way[k],treelayer[k]);
  76. }
  77. else if(type=='Q')
  78. {
  79. for(i=2;i<82;i++)
  80. ch2[i-2]=ch[i];
  81.  
  82. copy[0]=atof(strtok(ch2," \n\r\t"));
  83. copy[1]=atof(strtok(NULL," \n\r\t"));
  84. k=1;
  85. while(!(copy[0]==jws[k][0] && copy[1]==jws[k][1]))
  86. {
  87. if((jws[k][0]=='\0' && jws[k][1]=='\0')||(copy[0]=='\0' && copy[1]=='\0')){
  88. printf("NO\n");
  89. break;
  90. }
  91. if(copy[0]<=jws[k][0] && copy[1]<=jws[k][1])
  92. {
  93. k=4*k-2;
  94. }
  95. else if(copy[0]<=jws[k][0] && copy[1]>jws[k][1])
  96. {
  97. k=4*k-1;
  98. }
  99. else if(copy[0]>jws[k][0] && copy[1]<=jws[k][1])
  100. {
  101. k=4*k;
  102. }
  103. else if(copy[0]>jws[k][0] && copy[1]>jws[k][1])
  104. {
  105. k=4*k+1;
  106. }
  107. }
  108. if(copy[0]==jws[k][0] && copy[1]==jws[k][1])
  109. printf("%s %d %.1f %d\n",way[k],human[k],jws[k][2],treelayer[k]);
  110. }
  111. else if(type=='E')
  112. {
  113. printf("BYE\n");
  114. }
  115. }
  116. return 0;
  117. }
Add Comment
Please, Sign In to add comment