# poj 1011

Sticks

 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 122529 Accepted: 28391

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```

Source

Central Europe 1995

```#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
const int maxn=80;

char g[maxn][maxn];

bool vis[maxn][maxn];

int dis[maxn][maxn];

int go[4][2]={{-1,0},{1,0},{0,-1},{0,1}};

int x1,y1,x2,y2;

int w,h;

void bfs(int x,int y)
{
queue <int> q;

int tmp=x*(w+2)+y;

q.push(tmp);

vis[x][y]=true;

while(!q.empty())
{
tmp=q.front();

x=tmp/(w+2);

y=tmp%(w+2);

q.pop();

if(x==x2 && y==y2) return;

for(int i=0;i<4;i++)
{
int nx=x+go[i][0];
int ny=y+go[i][1];

while(nx>=0 && nx<=h+1 && ny>=0 && ny<=w+1 && !vis[nx][ny] && g[nx][ny]!=‘X‘)
{
tmp=nx*(w+2)+ny;
vis[nx][ny]=true;
dis[nx][ny]=dis[x][y]+1;
q.push(tmp);
nx=nx+go[i][0];
ny=ny+go[i][1];
}

}
}
}

int main()
{
int i;

int k=0;

while(scanf("%d%d",&w,&h)!=EOF && (w||h))
{
memset(g,0,sizeof(g));

for(i=1;i<=h;i++)
{
getchar();

for(int j=1;j<=w;j++)
scanf("%c",&g[i][j]);
}

printf("Board #%d:\n",++k);

int t=0;

while(scanf("%d%d%d%d",&y1,&x1,&y2,&x2) && (x1||y1||x2||y2))
{
memset(vis,false,sizeof(vis));

memset(dis,0,sizeof(dis));

g[x2][y2]=0;

bfs(x1,y1);

if(dis[x2][y2]==0)
printf("Pair %d: impossible.\n",++t);
else
printf("Pair %d: %d segments.\n",++t,dis[x2][y2]);

g[x2][y2]=‘X‘;    //注意还原
}

printf("\n");
}
return 0;
}
```

## POJ 1011 - Sticks DFS+剪枝

POJ 1011 - Sticks 题意:    一把等长的木段被随机砍成 n 条小木条    已知他们各自的长度,问原来这些木段可能的最小长度是多少 分析:    1. 该长度必能被总长整除    2. 从大到小枚举,因为小长度更灵活, 可拼接可不拼接    3. 因为每一跟木条都要用到, 故若轮到其中一根原始木段选它的第一根木条时,若第一根木条若不满足,则显然第一根木条在接下来任何一根原始木段都不会满足,故无解    4. 由于所有棒子已排序,在DFS时,若某根棒子未被选,则跳过其后面所有与

## [POJ 1011]Sticks(DFS剪枝)

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 original

## POJ 1011 Sticks dfs,剪枝 难度:2

http://poj.org/problem?id=1011 要把所给的集合分成几个集合,每个集合相加之和ans相等,且ans最小,因为这个和ans只在[1,64*50]内,所以可以用dfs一试 首先ans需要满足两个条件1.可以被总集合的和sum整除 2.是总集合的某个子集的和 对于条件1,可以通过试除遍历 对于条件2,可以通过dp预筛选,这两个花费时间都不大 接着搜索能不能划分成集合之和恰为ans的若干集合, 1. 可以从大向小找,因为大的更不灵活,而且所有的元素都需要取到 2.比如对于5,

## POJ 1011 sticks 搜索

Sticks Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 125918   Accepted: 29372 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 or

## POJ 1011 Sticks

DFS 题意是让你把这些木棍组合成长度相等的一些木棍.要求长度最短. #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int a[65],n,sum,ans; bool v[65],ok; bool cmpa(int a,int b) { return a>b; } bool dfs(int num,int need,