HDU 1709 The Balance
The Balance
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Problem Description
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.
Input
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.
Output
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.
Sample Input
3
1 2 4
3
9 2 1
Sample Output
0
2
4 5
Source
Recommend
给出一个天平兮,还有一些砝码。
给出一些质量兮,问你能称几个?
bitset水水,套个生成函数?
1 #include <cstdio> 2 #include <bitset> 3 4 std::bitset<100005> S; 5 6 signed main(void) 7 { 8 int n, m; 9 10 while (scanf("%d", &n) != EOF) 11 { 12 S.reset(), S.set(50000); 13 14 int ans = 0, sum = 0; 15 16 while (n--) 17 scanf("%d", &m), sum += m, 18 S = (S << m) | (S >> m) | S; 19 20 S >>= 50000; 21 22 for (int i = 1; i <= sum; ++i) 23 if (!S[i])++ans; 24 25 printf("%d\n", ans); 26 27 if (ans) 28 { 29 bool flag = false; 30 31 for (int i = 1; i <= sum; ++i) 32 if (!S[i]) 33 { 34 if (flag) 35 printf(" %d", i); 36 else 37 printf("%d", i), flag = true; 38 } 39 40 puts(""); 41 } 42 } 43 }
@Author: YouSiki