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标题:求解这两题T_T
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草狼
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求解这两题T_T
[Description]
E.T. Inc. employs Maryanna as alien signal researcher. To identify possible
alien signals and background noise, she develops a method to evaluate the signals
she has already received. The signal sent by E.T is more likely regularly
alternative.
Received signals can be presented by a string of small latin letters 'a' to 'z'
whose length is N. For each X between 1 and N inclusive, she wants you to find out
the maximum length of the substring which can be written as a concatenation of X
same strings. For clarification, a substring is a consecutive part of the original
string.
[Input]
The first line contains T, the number of test cases (T <= 200). Most of the test
cases are relatively small. T lines follow, each contains a string of only small
latin letters 'a' - 'z', whose length N is less than 1000, without any leading or
trailing whitespaces.
[Output]
For each test case, output a single line, which should begin with the case number
counting from 1, followed by N integers. The X-th (1-based) of them should be the
maximum length of the substring which can be written as a concatenation of X same
strings. If that substring doesn't exist, output 0 instead. See the sample for more
format details.

Sample Input  
2
arisetocrat
noonnoonnoon

Sample Output
Case #1: 11 0 0 0 0 0 0 0 0 0 0
Case #2: 12 8 12 0 0 0 0 0 0 0 0 0
[Hint]
For the second sample, the longest substring which can be written as a
concatenation of 2 same strings is "noonnoon", "oonnoonn", "onnoonno", "nnoonnoo",
any of those has length 8; the longest substring which can be written as a
concatenation of 3 same strings is the string itself. As a result, the second integer
in the answer is 8 and the third integer in the answer is 12.


(这题是区间DP,可是不会T_T)
[Description]
  Jerry loses himself in the interesting game: Fruit Ninja. Fruit Ninja is a game
of iPhone and iPad in which the players cut the fruits coming from the bottom of
the screen and gain the bonus from cutting more than two fruits with a single slice.
Once a fruit is cut, it breaks into small pieces and cannot be cut any more.
After months of training, he becomes pro of this game. Actually, he can cut all
the fruits on the screen at any time. Jerry also has a bad habit that he has no willing
to leave some fruits for the future cutting. In the other words, after Jerry cuts
the fruits, all the fruits on the screen breaks and no one left. That is why all
his friends call him “Juice Extractor”.
Now he only consider about the bonus, when he cuts more than two fruits, he can
gain some bonus scores as same as the number of fruits he slice at that time. For
example, if Jerry cuts 4 fruits with a single slice, he can get 4 scores from this
slice.
After Jerry gets the fruit schedule, he knows the appearing time and the
disappearing time for every single fruit. He can only cut a fruit into pieces between
its appearing time and disappearing time inclusive. He wants to know the maximum
possible bonus scores he can receive.
 
[Input]
  There are several test cases; the first line of the input contains a single
integer T, denoting the number of the test cases. (T <= 200)
For each test case, the first line contains an integer N, denoting the total
number of fruits. (1 <= N <= 1000)
The next N lines, each line describe a fruit. For each line, there are two
integers Xi and Yi, where Xi is the appearing time of the fruit and Yi is the
disappearing time of this fruit. (0 <= Xi <= Yi <= 1000000000)
 
[Output]
  For each test case, output a single integer denoting the maximum scores that
Jerry could possibly gain. See the sample for further details.
 
Sample Input
1
10
1 10
2 11
3 12
4 13
13 14

Sample Output
Case #1: 10

求思想,求高手指导,求交流,求各种求 emil:clddn3581927@
搜索更多相关主题的帖子: likely whose 
2011-05-17 21:43
waterstar
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狼兄我帮你顶顶

冰冻三尺,非一日之寒;士别三日,不足刮目相看!
2011-05-17 22:11
草狼
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回复 2楼 waterstar
非常感谢
2011-05-17 22:30
king_kong
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又看不懂     这是啥语言啊
2011-05-17 22:49
waterstar
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回复 3楼 草狼
狼兄,看不懂啊,翻译之后都看不懂,你上哪找来这么恶心的题?

冰冻三尺,非一日之寒;士别三日,不足刮目相看!
2011-05-18 12:21
linw1225
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帮你翻译下:

[说明]
东部公司的员工为外来信号研究员玛丽安娜。 为了确定可能
外来信号和背景噪声,她开发了一个方法来评估信号
她已经收到。 由内皮素发出的信号更容易定期
选择。
接收到的信号可以被提交一串小拉丁文字母'A'到'z'
它的长度为N之间每1和N包容十,她要你找出
子字符串的最大长度可作为书面的X串联
相同的字符串。 需要澄清,一子是一个连续的部分原
字符串。
[输入]
第一行包含T中,测试用例的数量(吨<= 200)。 大部分的测试
案件都比较小。 T线跟踪,每个包含一个唯一的小串
拉丁文字母'A' - 'Z'的,它的长度N小于1000,没有任何领导或
尾随空格。
[输出]
对于每一个测试案例,输出一行,应首先案件编号
从1数,由N个整数之后。 的X次(1型),其中应
子字符串的最大长度可作为连接的X同样的书面
字符串。 如果这样子不存在,输出0代替。 有关更多的样本
格式的细节。

采样输入  
2
arisetocrat
noonnoonnoon

输出范例
案例一:11 0 0 0 0 0 0 0 0 0 0
案例二:12 8 12 0 0 0 0 0 0 0 0 0
[提示]
对于第二个样本,最长的子串,可作为书面
2相同的字符串串联是“noonnoon”,“oonnoonn”,“onnoonno”,“nnoonnoo”
任何这些长度为8,最长的子串,可以作为书面
3相同的字符串串联是字符串本身。 因此,第二个整数
在答案是8,在回答第三个整数是12。


(这题是区间的DP,可是不会T_T)
[说明]
  杰里失去自己在有趣的游戏:水果忍者。 水果忍者是一个游戏
对iPhone和iPad,其中球员切水果从底部的到来
屏幕并得到一个片切割两个以上的水果奖金。
一旦果实被切断,它分解成小块,不能被削减了。
经过几个月的训练,他成为这个游戏的职业球员。 其实,他可以切断所有
在任何时候对屏幕上的成果。 杰里也有一个坏习惯,他没有愿意
留给未来的切割一些水果。 在换言之,削减后杰里
水果,所有的水果在屏幕上休息,没有人离开。 这就是为什么所有
他的朋友称他为“榨汁机”。
现在,他只考虑有关奖金,当他削减两个以上的水果,他可以
获得相同数量的水果片,他当时有些加分。 对于
例如,如果削减4杰里一个切片水果,他可以从中得到4分
片。
杰里取得成果后的时间表,他知道出现的时间和
消失的每一个果子的时候。 他只能减少一成块之间的水果
其出现的时间和消失时间的包容性。 他想知道最大
可以加分,他可以接受。

[输入]
  有几个测试用例,输入第一行包含一个单一
整数t,表示测试的个案。 (吨<= 200)
对于每个测试案例中,第一行包含一个整数N,表示总
水果的数量。 (1 <= N的<= 1000)
接下来的N行,每行描述一个水果。 对于每一行,有两个
整数朱熹与彝族,其中xi是水果及易出现的时间是
这果子的时候消失。 (0“=羲<=易<= 1000000000)

[输出]
  对于每一个测试案例,输出一个整数,意指的最高得分
杰里能可能收益。 进一步详情请参阅样本。

采样输入

10
1 10
2 11
3 12
4 13
13 14

输出范例
案例一:10

Einmal ist keinmal
2011-05-18 14:18
草狼
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回复 5楼 waterstar
我们复旦邀请赛的题目
 第一题的意思大概是这样的 给你一串字符串 让你找出一个最长子串,能把他平分成一样的X份(X=1,2,3,.....字符串总长)
如第2个案例: noonnoonnoon 分成一份的最长子串是其本身所以最长12,
              分成2分的最长子串有 oonnoonn,(oonn,oonn)或noonnoon(noon,noon)......最长为8;
             分成3分的最长子串是noonnoonnoon(noon,noon,noon) 最长为12;
             分成4,5,....12份得最长子串不存在所以长度为0;

第2题的意思大概是这样的 给你一堆水果,告诉你每个水果出现在屏幕上的开始时间和消失时间
                        让你去切水果,切到2个或2个以上就得相应的分数,一个不得分,
                        其中切一刀以后屏幕上的水果会全部消失,问你最多能得几分
2011-05-18 15:44
草狼
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终于把第一题做出来了,看到AC我泪流满面了
AC代码:

#include<iostream>
#include<cstring>
using namespace std;

int next[1009];
void getnext(char *str){
  int i=0;
  int j=-1;
  next[0]=-1;
  int len=strlen(str);
  while(i<len){
    if(j==-1|| str[i]==str[j] ){
      ++i;
      ++j;
      next[i]=j;
    }else j=next[j];
  }
}

int main(){
  int n;
  while(cin >> n){
    for(int cas=1; cas<=n; ++cas ){
      char str[1009];
      cin >> str;
      int sum[1009]={0};
      int len=strlen(str);
      for(int i=0; i<len; ++i ){
        getnext(str+i);
        int len1=strlen(str+i);
        for(int j=1; j<=len1; ++j){
          int ans=j-next[j];
          if(j%ans==0 && ans!=j){
            int tem=j/ans;
            int k=1;
            while(k<=tem){
              if(tem%k==0 && sum[k]<ans*tem) sum[k]=ans*tem;
              else if(sum[k]<ans*k) sum[k]=ans*k;
              ++k;
            }
          }
        }
      }
      sum[1]=len;
      cout << "Case #" << cas << ": ";
      for(int j=1; j<=len; ++j){
        cout << sum[j];
        if(j<len) cout << " ";
      }
      cout << endl;
    }
  }
  return 0;
}
2011-05-18 18:19
laoyang103
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帮顶

                                         
===========深入<----------------->浅出============
2011-05-18 18:31
我菜119
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回复 8楼 草狼
你参加的这个是ACM吗?

愿用余生致力编程
2011-05-18 18:43
快速回复:求解这两题T_T
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