Nyoj 293 sticks

SticksTime Limit: 3000 MS | memory limit: 65535 kb difficulty: 5

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 shoshould contains the smallest possible length of original sticks, one per line.

Sample Input

95 2 1 5 2 1 5 2 141 2 3 40

Sample output

65

Source

Poj

Uploaded

Zhang yuncong

Solution: Search + pruning... This is much harder than poj 1011 on nyoj. First, attach the poj 1011 AC code and paste the nyoj 293 AC code.

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long10 using namespace std;11 int d[70],n,ans,sum;12 bool vis[70];13 bool cmp(const int &a,const int &b) {14     return b < a;15 }16 bool dfs(int cur,int m,int len,int p) {17     if(m*len == sum-len) return true;18     if(cur == len) return dfs(0,m+1,len,0);19     for(int i = p; i < n; i++) {20         if(!vis[i] && cur+d[i] <= len) {21             vis[i] = true;22             if(dfs(cur+d[i],m,len,i+1)) return true;23             vis[i] = false;24             if(p == 0) return false;25             while(i+1 < n && d[i+1] == d[i]) i++;26         }27     }28     return false;29 }30 int main() {31     while(scanf("%d",&n),n) {32         int i;33         for(sum = i = 0; i < n; i++) {34             scanf("%d",d+i);35             sum += d[i];36         }37         ans = sum;38         sort(d,d+n,cmp);39         memset(vis,false,sizeof(vis));40         for(i = n; i > 1; i--)41             if(sum%i == 0 && dfs(0,0,sum/i,0)) {ans = sum/i;break;}42         printf("%d\n",ans);43     }44     return 0;45 }

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Nyoj 293... Haha. If the current stick has reached the expected length, but the following cannot piece together the expected length, you can only blame the current stick, so return immediately... Because the current length has been satisfied, continuing the loop will only lead to equal or shorter than the current length .... Return 0 illustrates the irrationality of the current combination of sticks.

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cstdlib> 5 # include <vector> 6 # include <climits> 7 # include <algorithm> 8 # include <cmath> 9 # define ll long long10 using namespace STD; 11 int d [70], n, ANS, sum; 12 bool vis [70]; 13 bool CMP (const Int & A, const Int & B) {14 return B <A; 15} 16 bool DFS (INT cur, int M, int Len, int p) {17 if (M * Len = sum-len) return true; 18 for (INT I = P; I <n; I ++) {19 if (! Vis [I] & cur + d [I] <= Len) {20 vis [I] = true; 21 if (cur + d [I] = Len) {22 if (DFS (0, m + 1, Len, 0) return true; 23 vis [I] = false; 24 return 0; // more optimization conditions than those on poj 25} else {26 if (DFS (cur + d [I], M, Len, I + 1) return true; 27 vis [I] = false; 28 If (cur = 0) return false; 29 While (I + 1 <n & D [I] = d [I + 1]) I ++; 30} 31} 32} 33 return false; 34} 35 int main () {36 while (scanf ("% d", & N), n) {37 int I; 38 memset (D, 0, sizeof (d); 39 for (sum = I = 0; I <n; I ++) {40 scanf ("% d", D + I ); 41 sum + = d [I]; 42} 43 ans = sum; 44 sort (D, D + N, CMP); 45 memset (VIS, false, sizeof (VIS); 46 for (I = N; I> 1; I --) {47 If (sum % I = 0 & DFS (0, 0, sum/I, 0) {ans = sum/I; break;} 48} 49 printf ("% d \ n", ANS); 50} 51 return 0; 52}

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