NYOJ 293 Sticks

Sticks

时间限制： 3000 ms  |  内存限制：65535 KB

难度： 5

 

描述

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.

 

输入

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

The output should contains the smallest possible length of original sticks, one per line.

样例输入

9

5 2 1 5 2 1 5 2 1

4

1 2 3 4

0

样例输出

6

5

来源

POJ

上传者

张云聪

解题：搜索+剪枝。。。NYOJ上这题比POJ 1011要难得多啊，先附上POJ 1011 AC代码 ，待会贴NYOJ 293 AC代码

 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 long

10 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 }

View Code

 

NYOJ　293.。。哈..如果当前棍子已经达到预期的长度，却导致后面的无法拼凑出预期的长度，只能怪当前的棍子，所以立即返回。。。因为当前已经满足这个长度了，继续循环，只会导致与当前长度相等或者变短。。。。return 0说明了当前棍子的组合的不合理性

 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 long

10 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;//比POJ上多的优化条件

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 }

View Code