C language Problem 201703_guess

 

 

 

A complete explanation of the original question is as follows:

Description

A competition was just over. It had 3 problems and n players. Eachplayer had an ID number from 1 to n.? Nal rank wasdecided by the total score of the 3 problems. the higher the total score was, the higher a player ranked (the smaller the rank number ). if two players gotthe same total score, the one with the smaller ID number got a higher rank. we 've known for each problem, how much score each player might get if he didn 'tsolve totally wrong (if solved totally wrong, the player got zero in theproblem ). However, we don't know whether a player did get score in a problem. For a predicted? Nal rank, you need to judge if the rank is possible.

Input

Input contains several cases. For each case,? Rst line is an integer n, (n <= 16384) to indicate the number ofplayers, followed by n lines, the ith of which contains three real numbers a, B, c (0 <= a, B, c <1000. a, B and c have 2 decimal places at most .) torespectively indicate the score of each problem Player I might get if he didn 'tsolve totally wrong. another line containing n integers follows to indicate theplayer ID number in the order from rank 1st to rank nth.

The last caseis followed by a line containing only a zero.

Output

No solution "(quotes for clarity ).

Sample Input

3

100 200 300

100 200 300

100 200 300

1 2 3

3

100 200 300

100 200 300

100 200 300

3 2 1

0

Sample Output

Case 1: 600.00

Case 2: 400.00

 

The source of this question is ACM Beijing 2006-E.

I wrote the code myself and submitted it to the AC. The Code is as follows:

1. # include <stdio. h> 2. # include <stdlib. h> 3. 4. # define PROBLEM_NUM 3 5. 6. int main () 7. {8. int n = 0; // number of players 9. float ** grades = NULL; // possible grades if not totally wrong 10. float results [1000]; 11. int * rank = NULL; // predicted rankings 12. float grade = 0, temp = 0, temp1 = 0; 13. int m = 0; // case counter: case m 14. int I = 0, j = 0, k = 0; // loop counter 15. 16. s Canf ("% d", & n); 17. while (n) {18. rank = (int *) malloc (sizeof (int) * n); 19. grades = (float **) malloc (sizeof (float *) * n); 20. for (I = 0; I <n; I ++) 21. grades [I] = (float *) malloc (sizeof (float) * PROBLEM_NUM); 22. 23. // get input 24. for (I = 0; I <n; I ++) 25. for (j = 0; j <PROBLEM_NUM; j ++) {26. scanf ("% f", & temp); 27. getchar (); // get rid of white spaces 28. /* Insertion Sort */29. for (K = j; k> 0 & temp <grades [I] [k]; k --) 30. grades [I] [k] = grades [I] [k-1]; 31. grades [I] [k] = temp; 32 .} 33. 34. for (I = 0; I <n; getchar (), I ++) 35. scanf ("% d", & rank [I]); 36. 37. // judge the rank and find the score 38. grade = grades [rank [0]-1] [0] + grades [rank [0]-1] [1] + grades [rank [0]-1] [2]; 39. temp = 0; 40. for (I = 1; I <n; I ++) {41. if (rank [I]> rank [I-1]) {42. for (j = 0; j <8 & temp <= grade; j ++) {43. switch (j) {// value of "temp" increases with "j" 44. case 0: temp = 0; break; 45. case 1: temp = grades [I] [0]; break; 46. case 2: temp = grades [I] [1]; break; 47. case 3: 48. temp = grades [I] [0] + grades [I] [1]> = grades [I] [2]? Grades [I] [2]: grades [I] [0] + grades [I] [1]; 49. break; 50. case 4: 51. temp = grades [I] [0] + grades [I] [1] <= grades [I] [2]? Grades [I] [2]: grades [I] [0] + grades [I] [1]; 52. break; 53. case 5: temp = grades [I] [0] + grades [I] [2]; break; 54. case 6: temp = grades [I] [1] + grades [I] [2]; break; 55. case 7: temp = grades [I] [0] + grades [I] [1] + grades [I] [2]; break; 56. default: break; 57 .} 58. temp1 = temp <= grade? Temp: temp1; 59 .} 60 .} 61. else {62. for (j = 0; j <8 & temp <grade; j ++) {63. switch (j) {// value of "temp" increases with "j" 64. case 0: temp = 0; break; 65. case 1: temp = grades [I] [0]; break; 66. case 2: temp = grades [I] [1]; break; 67. case 3: 68. temp = grades [I] [0] + grades [I] [1]> = grades [I] [2]? Grades [I] [2]: grades [I] [0] + grades [I] [1]; 69. break; 70. case 4: 71. temp = grades [I] [0] + grades [I] [1] <= grades [I] [2]? Grades [I] [2]: grades [I] [0] + grades [I] [1]; 72. break; 73. case 5: temp = grades [I] [0] + grades [I] [2]; break; 74. case 6: temp = grades [I] [1] + grades [I] [2]; break; 75. case 7: temp = grades [I] [0] + grades [I] [1] + grades [I] [2]; break; 76. default: break; 77 .} 78. temp1 = temp <grade? Temp: temp1; 79 .} 80 .} 81. 82. if (j = 1) break; 83. 84. grade = temp1; 85. temp = temp1 = 0; 86 .} 87. if (I = n & j! = 1) 88. results [m] = grade; 89. else 90. results [m] =-1; 91. m ++; 92. 93. // release storage 94. for (I = 0; I <n; I ++) 95. free (grades [I]); 96. free (grades); 97. free (rank); 98. 99. scanf ("% d", & n); // player number of next case 100 .} 101. 102. for (I = 0; I <m; I ++) {103. if (results [I]! (=-1) 104. printf ("Case % d: %. 2f \ n ", I, results [I]); 105. else 106. printf ("Case % d: No solution \ n"); 107 .} 108. // free (results); 109. 110. return 0; 111 .}

My idea is to sort the inserts when reading data, and then use the switch... case... statement to try all possible conditions starting from the minimum value.

 

This code is not very well handled in many places.

1. A fixed-length array (large enough) results [1000] is used to store the results.

2. the number of problems cannot be easily changed. When PROBLEM_NUM is not 3, this Code cannot be used directly; this is mainly restricted by the switch in the judgment part... case... statement. The method I use is to try from the minimum possible score until the maximum possible, a little poor taste, I think there will be a better algorithm.

3. Almost the same code is repeated, that is, the two switches... case... statements.

You are welcome to give your thoughts and comments. Let me learn ~

This article is from the "WhatWhyHow" blog