2018 codejam kickstart H轮

被第一题傻逼题卡了很久……好的我也是个傻逼

倒在了最后一题本来以为小数据过了就能过大数据结果下载了大数据发现怎么输出了好多个零调代码过程中超时了结束后重新提交了一下果然是不通过的

题目

Problem

You are a contestant on a popular new game show and are playing for the grand prize!

There are two big buttons, a red one and a black one. You will make a sequence of exactly N button presses.

There are lots of different sequences of presses you could make, but there are Pforbidden prefixes, each of length no greater than N. If you make a sequence of presses which begins with any of the forbidden sequences, you will not win the grand prize. It is fine for your sequence to contain one or more forbidden prefixes as long as they do not appear at the start of your sequence.

A winning sequence must consist of exactly N button presses and must not begin with one of the forbidden prefixes. How many different winning sequences are there?

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line containing two integers N and P, as described above. Then, there are P more lines, each of which contains a string of between 1 and N characters, inclusive, describing one of the forbidden sequences of presses. An R represents pressing the red button, whereas a B represents pressing the black button.

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the number of winning sequences, as desribed above.

Limits

1 ≤ T ≤ 100.
1 ≤ P ≤ min(2^N, 100).
Each forbidden prefix is between 1 and N characters long, inclusive.
No two forbidden prefixes will be the same.

Small dataset

1 ≤ N ≤ 10.

Large dataset

1 ≤ N ≤ 50.

Sample

Input	Output
4 3 2 BBB RB 5 1 R 4 3 R B RBRB 50 5 BRBRBBBRBRRRBBB BRBRBRRRBRRRBRB BBBRBBBRBRRRBBB BRBRBRRRBRRRB BRBRBBBRBBBRB	Case #1: 5 Case #2: 16 Case #3: 0 Case #4: 1125556309458944

Note that the last Sample case would not appear in the Small dataset.

In the first case, you must make a sequence of 3 presses. There are 8 possible sequences of three presses, but some of them will cause you to lose the game. They are listed below:

RBB. This is forbidden since it starts with the first forbidden sequence (RB).
RBR. This is forbidden since it starts with the first forbidden sequence (RB).
BBB. This is forbidden since it starts with the second forbidden sequence (BBB).

Thus, there are only 5 winning sequences.

In the second case, you must make a sequence of 5 presses. There is only one forbidden sequence, which is R. This means that the first press must be B, and the next 4 presses can be either button. This gives a total of 16 different button presses.

In the third case, you must make a sequence of 4 presses. There are three forbidden sequences, but since every possible sequence begins with either R (the first forbidden sequence) or B (the second forbidden sequence), there are no winning sequences. So the answer is 0.

题意

　　　给出一个仅存在R和B的长度为n的字符串，有m个不能出现的前缀，问这样的字符串有多少。

思路

　　　崔神给我讲的思路是可以用字典树先对m个前缀建树，然后用dfs剪枝。

　　　我自己的做法是对前缀暴力找是否存在一个前缀为另一个前缀的前缀（前缀 i 为前缀 j 的前缀）。若存在，则将长度更长的前缀标记，计算时不再计入。

　　　对于一个前缀 s 而言，剩余长度（n-s.length()）无论怎么排列R和B都是不合法的。

　　　对计入计算的前缀 s[i] ，累加（2^(n-s[i].length())，最后用2^n减去累加结果。

代码（大小数据均为同一份代码，只改动了文件关联）

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <string>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int inf=0x3f3f3f3f;

const char ss[]={'B','R'};

int t,n,m,vis[];

string s[];

bool cmp(string s1,string s2){

    if(s1.length()<s2.length()) return s1.length()<s2.length();

}

ll qpow(ll x,ll k){

    ll res=;

    while(k>){

       if(k%==){

         res=res*x;

       }

       x=x*x;

       k>>=;

    }

    return res;

}  

int main(){

//    freopen("A-large.in","r",stdin);

//    freopen("A.out","w",stdout);

    scanf("%d",&t);

    for(int id=;id<=t;id++){

        memset(vis,,sizeof(vis));

        scanf("%d%d",&n,&m);

        for(int i=;i<=m;i++){

            cin>>s[i];

        }

        sort(s+,s+m+,cmp);

        for(int i=;i<=m;i++){

            for(int j=i+;j<=m;j++){

                if(vis[i]== || vis[j]==) continue;

                int flag=;

                for(int k=;k<s[i].length();k++){

                    if(s[i][k]==s[j][k]) flag++;

                    if(flag==s[i].length()) vis[j]=;

                }

            }

        }

        ll ans=qpow(,n);

        for(int i=;i<=m;i++){

            if(vis[i]==){

                ans-=qpow(,n-s[i].length());

            }

        }

        printf("Case #%d: %lld\n",id,ans);

    }

    return ;

}

题目

Problem

Thanh wants to paint a wonderful mural on a wall that is N sections long. Each section of the wall has a beauty score, which indicates how beautiful it will look if it is painted. Unfortunately, the wall is starting to crumble due to a recent flood, so he will need to work fast!

At the beginning of each day, Thanh will paint one of the sections of the wall. On the first day, he is free to paint any section he likes. On each subsequent day, he must paint a new section that is next to a section he has already painted, since he does not want to split up the mural.

At the end of each day, one section of the wall will be destroyed. It is always a section of wall that is adjacent to only one other section and is unpainted (Thanh is using a waterproof paint, so painted sections can't be destroyed).

The total beauty of Thanh's mural will be equal to the sum of the beauty scores of the sections he has painted. Thanh would like to guarantee that, no matter how the wall is destroyed, he can still achieve a total beauty of at least B. What's the maximum value of B for which he can make this guarantee?

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line containing an integer N. Then, another line follows containing a string of N digits from 0 to 9. The i-th digit represents the beauty score of the i-th section of the wall.

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the maximum beauty score that Thanh can guarantee that he can achieve, as described above.

Limits

1 ≤ T ≤ 100.

Small dataset

2 ≤ N ≤ 100.

Large dataset

For exactly 1 case, N = 5 × 10⁶; for the other T - 1 cases, 2 ≤ N ≤ 100.

Sample

Input	Output
4 4 1332 4 9583 3 616 10 1029384756	Case #1: 6 Case #2: 14 Case #3: 7 Case #4: 31

In the first sample case, Thanh can get a total beauty of 6, no matter how the wall is destroyed. On the first day, he can paint either section of wall with beauty score 3. At the end of the day, either the 1st section or the 4th section will be destroyed, but it does not matter which one. On the second day, he can paint the other section with beauty score 3.

In the second sample case, Thanh can get a total beauty of 14, by painting the leftmost section of wall (with beauty score 9). The only section of wall that can be destroyed is the rightmost one, since the leftmost one is painted. On the second day, he can paint the second leftmost section with beauty score 5. Then the last unpainted section of wall on the right is destroyed. Note that on the second day, Thanh cannot choose to paint the third section of wall (with beauty score 8), since it is not adjacent to any other painted sections.

In the third sample case, Thanh can get a total beauty of 7. He begins by painting the section in the middle (with beauty score 1). Whichever section is destroyed at the end of the day, he can paint the remaining wall at the start of the second day.

题意

　　　有一个由 0~9 的数字组成的数列，第一次可以随便取一个数字，接下来的每一次只能取一个与已取区间相邻的数字。

　　　每次取出一个数字后，数列两端的两个数之一会随机地变为不可取。问最坏情况下取出数字的最大和是多少。

思路

　　　首先知道我们一定能取出一个长度为 (n+1)/2的连续区间；
　　　而且，不管不可取的数字是如何出现在两端的，对于一个长度为 (n+1)/2 的区间，一定存在一种方法把它取到。
　　　所以 O(n) 找出区间和最大的长度为 (n+1)/2 的连续区间即可。

代码

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <string>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int inf=0x3f3f3f3f;

const int maxn=5e6+;

int t,n;

int a[maxn],sum[maxn];

char s[maxn];

int main(){

//    freopen("B-large.in","r",stdin);

//    freopen("B.out","w",stdout);

    scanf("%d",&t);

    for(int id=;id<=t;id++){

        memset(sum,,sizeof(sum));

        scanf("%d",&n);

        scanf("%s",s+);

        for(int i=;i<=n;i++){

            a[i]=s[i]-'';

        }

        sum[]=;

        for(int i=;i<=n;i++){

            sum[i]=sum[i-]+a[i];

        }

        int m=(n+)/;

        int maxx=;

        for(int i=m;i<=n;i++){

            maxx=max(maxx,sum[i]-sum[i-m]);

        }

        printf("Case #%d: %d\n",id,maxx);

    }

    return ;

}

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