NC51014. Sticks
描述
输入描述
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.
输出描述
The output should contains the smallest possible length of original sticks, one per line.
示例1
输入:
9 5 2 1 5 2 1 5 2 1 4 1 2 3 4 0
输出:
6 5
C++14(g++5.4) 解法, 执行用时: 4ms, 内存消耗: 384K, 提交时间: 2020-07-19 17:57:00
#include <bits/stdc++.h> using namespace std; int a[100],vis[100],n,num,len; bool dfs(int sticks,int rest,int last) { if(sticks==num+1) return true; if(rest==0) return dfs(sticks+1,len,0); int fail=-1; for(int i=last+1;i<=n;i++) if(!vis[i]&&rest>=a[i]&&a[i]!=fail) { vis[i]=true; if(dfs(sticks,rest-a[i],i)) return true; vis[i]=false; fail=a[i]; if(rest==len||rest==a[i]) return false; } return false; } int main() { while(cin>>n&&n) { int sum=0,m=0,x; for(int i=1;i<=n;i++) { cin>>x; if(x<=50) {a[++m]=x;sum+=x;} } n=m; sort(a+1,a+n+1,greater<int>()); memset(vis,0,sizeof(vis)); for( len=a[1];len<=sum;len++) { if(sum%len) continue; num=sum/len; if(dfs(1,len,0)) break; } cout<<len<<endl; } return 0; }
C++(g++ 7.5.0) 解法, 执行用时: 5ms, 内存消耗: 404K, 提交时间: 2023-03-11 21:10:00
#include<bits/stdc++.h> using namespace std; const int N=110; int a[N],cnt,len,n; int v[N]; bool dfs(int p,int c,int lst) { if(p==cnt) return true; if(c==len) return dfs(p+1,0,0); int fl=0; for(int i=lst;i<n;++i){ if(!v[i]&&c+a[i]<=len&&fl!=a[i]){ v[i]=1; if(dfs(p,c+a[i],i+1)) return true; fl=a[i]; v[i]=0; if(c==0||c+a[i]==len) return false; } } return false; } int main() { while(cin>>n&&n){ int s=0; for(int i=0;i<n;++i){ cin>>a[i]; if(a[i]>50){ --i;--n; continue; } s+=a[i]; } sort(a,a+n); reverse(a,a+n); for(len=1;len<=s;++len){ if(s%len) continue; cnt=s/len; memset(v,0,sizeof v); if(dfs(0,0,0)) break; } cout<<len<<endl; } return 0; }
C++ 解法, 执行用时: 15ms, 内存消耗: 384K, 提交时间: 2021-06-15 21:58:51
#include<bits/stdc++.h> using namespace std; int a[110],v[110],n,tot,cnt,l; int dfs(int stick,int cab) { if(stick>cnt) return 1; if(cab==l) return dfs(stick+1,0); int fail=0; for(int i=1;i<=tot;i++) { if(!v[i]&&cab+a[i]<=l&&fail!=a[i]) { v[i]=1; if(dfs(stick,cab+a[i])) return 1; v[i]=0; fail=a[i]; if(cab==0||cab+a[i]==l) return 0; } } return 0; } int main() { while(~scanf("%d",&n),n) { tot=0;int val=0,s=0;memset(v,0,sizeof(v)); for(int i=1;i<=n;i++) { int x;scanf("%d",&x); if(x>50) continue; a[++tot]=x; s+=x; val=max(val,x); } sort(a+1,a+1+tot); reverse(a+1,a+1+tot); for( l=val;l<s;l++) { if(s%l) continue; cnt=s/l; if(dfs(1,0)) break; } printf("%d\n",l); } }