Problem Description
George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the original state, but he forgot how many sticks he had originally and how long they were originally. Please help him and design a program which computes the smallest possible original length of those sticks. All lengths expressed in units are integers greater than zero.
Input
The input contains blocks of 2 lines. The first line contains the number of sticks parts after cutting, there are at most 64 sticks. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.
Output
The output file contains the smallest possible length of original sticks, one per line.
SampleInput
9 5 2 1 5 2 1 5 2 1 4 1 2 3 4 0
SampleOutput
6 5
题意就是给你一堆不同长度的木棍，要你用他们拼成长度相等的木棍，输出最短且满足要求的木棍长度。
就直接从0开始搜嘛，判断长度相同就输出，好了就这样，提交，TLE。
=7=那么看来不能直接搜索，如何优化呢，当然就是判断一些条件剪枝，首先是搜索范围，根据题意有原始长度肯定比现在最长的要长，而且能被现在总长除尽，原始长度最长肯定就是总长度（只有一根木棍），其次就是对于同一个长度的木棍，只尝试一次，一次不成功便不需要尝试第二次，然后就依次dfs就行了。
代码：
1 #include <iostream> 2 #include <string> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <sstream> 6 #include <iomanip> 7 #include <map> 8 #include <stack> 9 #include <deque> 10 #include <queue> 11 #include <vector> 12 #include <set> 13 #include <list> 14 #include <cstring> 15 #include <cctype> 16 #include <algorithm> 17 #include <iterator> 18 #include <cmath> 19 #include <bitset> 20 #include <ctime> 21 #include <fstream> 22 #include <limits.h> 23 #include <numeric> 24 25 using namespace std; 26 27 #define F first 28 #define S second 29 #define mian main 30 #define ture true 31 32 #define MAXN 1000000+5 33 #define MOD 1000000007 34 #define PI (acos(-1.0)) 35 #define EPS 1e-6 36 #define MMT(s) memset(s, 0, sizeof s) 37 typedef unsigned long long ull; 38 typedef long long ll; 39 typedef double db; 40 typedef long double ldb; 41 typedef stringstream sstm; 42 const int INF = 0x3f3f3f3f; 43 44 int s[500050]; 45 bool vis[500050]; 46 int sum,n; 47 48 bool dfs(int pos, int res, int len){ 49 if(pos == 0 && res == 0) 50 return true; 51 if(!res) 52 res = len; 53 for(int i = n - 1; i >= 0; i--){ 54 if(vis[i] || s[i] > res) 55 continue; 56 vis[i] = true; 57 if(dfs(pos - 1, res - s[i], len)) 58 return true; 59 vis[i] = false; 60 if(res == s[i] || res == len) //判断是否尝试过 61 return false; 62 } 63 return false; 64 } 65 66 int main(){ 67 ios_base::sync_with_stdio(false); 68 cout.tie(0); 69 cin.tie(0); 70 while(cin >> n && n){ 71 sum = 0; 72 for(int i = 0; i < n; i++){ 73 cin >> s[i]; 74 sum += s[i]; 75 } 76 sort(s, s+n); 77 int i; 78 for(i = s[n - 1]; i <= sum/2;i++){ //从最大长度开始dfs 79 if(sum % i) //小于总长度则dfs尝试 80 continue; 81 MMT(vis); //每次dfs前记得清空标记 82 if(dfs(n,i,i)){ 83 cout << i << endl; 84 break; 85 } 86 } 87 if(i > sum/2) //大于长度一半，则不可能被分成多个木棍，肯定只能由一根木棍构成 88 cout << sum << endl; 89 } 90 return 0; 91 }
