1. #include <stdio.h>
2. #include <string.h>
3. #include <stdlib.h>
4. #define MAX 10010
5. int c1[MAX], c2[MAX], num[110];
6. /* 题目大意： 给你n种砝码，从1到n中砝码的重量和其中不能称出多少种重量，输出不能称出的总数和类别*/
7. int main()
8. {
9. int n;
10. while( scanf("%d", &n)!=EOF )
11. {
12. memset(num, 0, sizeof(num) );
13. memset(c1, 0, sizeof(c1) );
14. memset(c2, 0, sizeof(c2) );
15. int total = 0;
16. for(int i=0; i<n; i++) //重量输入
17. {scanf("%d", &num[i]); total+=num[i];}
18. //for(int i=0; i<=total; i++)
19. //c1[i] = c2[i] = 0;
20. c1[0] = 1; //第一个括号，重量为0的
21. for(int i=0; i<n; i++)
22. {
23. for(int j=0; j<=total; j++)
24. {
25. c2[j] |= c1[j]; //否能称得，自己砝码的重量
26. if(num[i] + j <=total) //重量和是否大于总重量
27. c2[j+num[i]] |= c1[j]; //重量和
28. c2[abs(j-num[i])] |= c1[j]; //重量差
29. }
30. for(int j=0; j<=total; j++) //一次循环赋值
31. {c1[j] = c2[j]; c2[j] = 0;}
32. }
33. int count = 0;
34. for(int i=0; i<=total; i++) //查找能否称得的重量
35. {
36. if(!c1[i])
37. c2[count++] = i;
38. }
39. printf("%d\n", count);
40. for(int i=0; i<count-1; i++)
41. printf("%d ", c2[i]);
42. if(count) printf("%d\n", c2[count-1]);
43. }
44. return 0;
45. }
46. /*
47. Problem Description
48. Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range [1,S]. S is the total quality of all the weights.
49. Input
50. The input consists of multiple test cases, and each case begins with a single positive integer N (1<=N<=100) on a line by itself indicating the number of weights you have. Followed by N integers Ai (1<=i<=N), indicating the quality of each weight where 1<=Ai<=100.
51. Output
52. For each input set, you should first print a line specifying the number of qualities which cannot be measured. Then print another line which consists all the irrealizable qualities if the number is not zero.
53. Sample Input
54. 3
55. 1 2 4
56. 3
57. 9 2 1
58. Sample Output
59. 0
60. 2
61. 4 5
62. */

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