漂亮小姐姐点击就送:http://poj.org/problem?id=1011
Sticks
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 151718 | Accepted: 36127 |
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 should contains the smallest possible length of original sticks, one per line.
Sample Input
9 5 2 1 5 2 1 5 2 1 4 1 2 3 4 0
Sample Output
6 5
//题意:求n个数随意组合成相等的数,组合起来的数最小是几 //emmmmm 就是,现在有一堆木棒,他们是被切割之后的,他们有长度,问他们被切割之前的可能的最小长度是多少 //思路:求出最大值、所有数的和,然后从小到大枚举 //把他们扔进vector里,在里边二分 //不断删数,知道vec为空 //lower_bound查找方向要和vector的方向一样。。 //lower_bound。。。。。。。害怕.jpg //我去竟然WA掉。。我以为会TLE掉的。。。。。 //气死。。写俩小时竟然不对。。。 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> using namespace std; const int N=70; int T,n,sum; vector<int> vec,v; inline int read() { char c=getchar();int num=0; for(;!isdigit(c);c=getchar()); for(;isdigit(c);c=getchar()) num=num*10+c-'0'; return num; } int main() { while(n=read()) { if(!n) break; sum=0; vec.clear(); for(int i=1,a;i<=n;++i) { a=read(); sum+=a; vec.push_back(a); } int now; sort(vec.begin(),vec.end(),greater<int>() ); for(int ans=vec[0];ans<=sum;++ans) { v.clear(); for(int i=0;i<n;++i) { if(vec[i]!=ans) v.push_back(vec[i]); } now=0; bool flag=1; while(v.size()) { // cout<<"Sz: "<<v.size()<<endl; vector<int>::iterator pos=lower_bound(v.begin(),v.end(),ans-now,greater<int>() ); if(pos==v.end()) { break; } else if((*pos)+now==ans) { v.erase(pos); if(v.size()) { now=v[0]; v.erase(v.begin()); } else now=-1; } else { now+=(*pos); v.erase(pos); } } if(now==-1) { printf("%d\n",ans); break; } } } return 0; }