Baudelaire is Programmer

When John was a little kid he didn't have much to do. There was no internet, no Facebook, and no programs to hack on. So he did the only thing he could... he evaluated the beauty of strings in a quest to discover the most beautiful string in the world.

Given a string s, little Johnny defined the beauty of the string as the sum of the beauty of the letters in it.

The beauty of each letter is an integer between 1 and 26, inclusive, and no two letters have the same beauty. Johnny doesn't care about whether letters are uppercase or lowercase, so that doesn't affect the beauty of a letter. (Uppercase 'F' is exactly as beautiful as lowercase 'f', for example.)

You're a student writing a report on the youth of this famous hacker. You found the string that Johnny considered most beautiful. What is the maximum possible beauty of this string?

Input

The input file consists of a single integer m followed by m lines.

Output

Your output should consist of, for each test case, a line containing the string "Case #x: y" where x is the case number (with 1 being the first case in the input file, 2 being the second, etc.) and y is the maximum beauty for that test case.

Constraints

5 ≤ m ≤ 50
2 ≤ length of s ≤ 500

贪心地把最大beauty给出现频率最大的letter即可。

#include<iostream>
#include<fstream>
#include<string.h>
#include<algorithm>
using namespace std;
//char s[501];
string s;
int m;
ifstream fin;
ofstream fout;
int main(){
	fin.open("beautiful_stringstxt.txt");
	fout.open("output.txt");
	fin>>m;
	getline(fin,s);
	for(int cas=1;cas<=m;cas++){
		getline(fin,s);
		//cout<<s<<endl;
		int len=s.size();
		int count[26];
		for(int i=0;i<26;i++)count[i]=0;
		for(int i=0;i<len;i++){
			char c=s[i];
			if(c<='Z'&&c>='A')c=c-'A'+'a';
			if(c<='z'&&c>='a'){
				count[c-'a']++;
			}
		}
		sort(count,count+26);
		int ans=0;
		for(int i=26;i>=1;i--){
			ans+=i*count[i-1];
		}
		fout<<"Case #"<<cas<<": "<<ans<<endl;
	}
	return 0;
}

Your friend John uses a lot of emoticons when you talk to him on Messenger. In addition to being a person who likes to express himself through emoticons, he hates unbalanced parenthesis so much that it makes him go :(

Sometimes he puts emoticons within parentheses, and you find it hard to tell if a parenthesis really is a parenthesis or part of an emoticon.

A message has balanced parentheses if it consists of one of the following:

• - An empty string ""
• - One or more of the following characters: 'a' to 'z', ' ' (a space) or ':' (a colon)
• - An open parenthesis '(', followed by a message with balanced parentheses, followed by a close parenthesis ')'.
• - A message with balanced parentheses followed by another message with balanced parentheses.
• - A smiley face ":)" or a frowny face ":("

Write a program that determines if there is a way to interpret his message while leaving the parentheses balanced.

Input

The first line of the input contains a number T (1 ≤ T ≤ 50), the number of test cases.
The following T lines each contain a message of length s that you got from John.

Output

For each of the test cases numbered in order from 1 to T, output "Case #i: " followed by a string stating whether or not it is possible that the message had balanced parentheses. If it is, the string should be "YES", else it should be "NO" (all quotes for clarity only)

Constraints

• 1 ≤ length of s ≤ 100

我写了一个O(n³)的DP，思路正确结果WA了，原因是忽略了“（）”这种边界情况：

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

int T;
string s;
ifstream fin;
ofstream fout;
bool dp[101][101];
int main(){
	fin.open("balanced_smileystxt.txt");
	fout.open("output.txt");
	fin>>T;
	getline(fin,s);
	for(int cas=1;cas<=T;cas++){
		for(int i=0;i<101;i++)for(int j=0;j<101;j++)dp[i][j]=false;
		getline(fin,s);
		int len=s.size();
		int k;
		for(k=0;k<len;k++){
			char c=s[k];
			if(c!=' '&&(c<'a'||c>'z')&&c!=':'&&c!='('&&c!=')'){
				break;
			}
		}
		if(k<len){
			fout<<"Case #"<<cas<<": NO"<<endl;
			continue;
		}
		if(len==0){
			fout<<"Case #"<<cas<<": YES"<<endl;
			continue;
		}
		for(int l=1;l<=len;l++){
			for(int i=0;i+l-1<len;i++){
				if(l==1){
					if(s[i]!='('&&s[i]!=')')dp[i][i]=true;
					else dp[i][i]=false;
					continue;
				}
				else if(l==2){//忽略了“()”的情况!!
					if((s[i]==':'&&s[i+1]==')')||(s[i]==':'&&s[i+1]=='(')){
						dp[i][i+1]=true;
					}
				}
				for(int k=i;k<i+l-1;k++){
					if(dp[i][k]&&dp[k+1][i+l-1]){
						dp[i][i+l-1]=true;
						break;
					}
				}
				if(l>=3){
					if(s[i]=='('&&s[i+l-1]==')'&&dp[i+1][i+l-2]){
						dp[i][i+l-1]=true;
					}
				}
			}
		}
		fout<<"Case #"<<cas<<": ";
		if(dp[0][len-1])fout<<"YES";
		else fout<<"NO";
		fout<<endl;
	}
	return 0;
}

官方题解给出了一个O(n)的答案，实际上这里只需要检测括号是否有可能匹配。

After sending smileys, John decided to play with arrays. Did you know that hackers enjoy playing with arrays? John has a zero-based index array, m, which contains n non-negative integers. However, only the first k values of the array are known to him, and he wants to figure out the rest.

John knows the following: for each index i, where k <= i < n, m[i] is the minimum non-negative integer which is *not* contained in the previous *k* values of m.

For example, if k = 3, n = 4 and the known values of m are [2, 3, 0], he can figure out that m[3] = 1.

John is very busy making the world more open and connected, as such, he doesn't have time to figure out the rest of the array. It is your task to help him.

Given the first k values of m, calculate the n^th value of this array. (i.e. m[n - 1]).

Because the values of n and k can be very large, we use a pseudo-random number generator to calculate the first k values of m. Given non-negative integers a, b, c and positive integer r, the known values of m can be calculated as follows:

• m[0] = a
• m[i] = (b * m[i - 1] + c) % r, 0 < i < k

Input

The first line contains an integer T (T <= 20), the number of test cases.
This is followed by T test cases, consisting of 2 lines each.
The first line of each test case contains 2 space separated integers, n, k (1 <= k <= 10⁵, k < n <= 10⁹).
The second line of each test case contains 4 space separated integers a, b, c, r (0 <= a, b, c <= 10⁹, 1 <= r <= 10⁹).

Output

For each test case, output a single line containing the case number and the n^th element of m.

注意到序列是循环的这个问题就很好解决了，只需要求出第一个“循环节”。这时候一个暴力的方法是O(k²)的，仍然难以接受，为此我加入了一个小优化：

考虑a,x₁,...,x_k,y这个序列片段，现在我们要求y，如果a<x_k并且a在x₁~x_k-1 中没有再次出现，则y=a；否则我们只需从小到大检验大于x_k的数。这样在最坏情况下也可以秒杀。

#include<iostream>
#include<fstream>

using namespace std;
int had[100001];
int circle[100001];
int memo[100001];
int T;
long long n,k;
long long a,b,c,r;
ifstream fin;
ofstream fout;

int main(){
    fin.open("find_the_mintxt.txt");
    fout.open("output.txt");
    fin>>T;
    for(int cas=1;cas<=T;cas++){
        fin>>n>>k;
        fin>>a>>b>>c>>r;
        
        for(int i=0;i<=k;i++)had[i]=0;
        
        long long pre=a;
        if(a<=k)had[a]++;
        memo[0]=a;
        for(int i=1;i<k;i++){
            pre=(b*pre+c)%r;
            if(pre<=k)had[pre]++;
            memo[i]=pre;
            //fout<<memo[i]<<' ';
        }
        //fout<<endl;
        int j;
        for(j=0;j<=k;j++){
            if(had[j]==0)break;
        }
        circle[0]=j;
        if(memo[0]<=k)had[memo[0]]--;
        had[j]++;
        //fout<<circle[0]<<' ';
        
        for(int i=1;i<=k;i++){
            
            if(memo[i-1]<circle[i-1]&&had[memo[i-1]]==0){
                circle[i]=memo[i-1];
            }
            else{
                int x;
                for(x=circle[i-1]+1;x<=k;x++){
                    if(had[x]==0)break;
                }
                circle[i]=x;
            }
            
            
            /*
            int j;
            for(j=0;j<=k;j++){
                if(had[j]==0)break;
            }
            circle[i]=j;
            */
            
            if(i<k){
                if(memo[i]<=k)had[memo[i]]--;
                had[circle[i]]++;
            }
            //fout<<circle[i]<<' ';
        }
        //fout<<endl;
        fout<<"Case #"<<cas<<": "<<circle[(n-k-1)%(k+1)]<<endl;
    }
    return 0;
}

官方题解：

from collections import Counter

def get_beauty(string):

    string = string.lower()

    # Remove all characters other than letters

    string = ''.join(x for x in string if 'a' <= x <= 'z' )

    # Make a dictionary where the keys are letters and the values are counts

    freq = Counter(string)

    # Get the values (letter counts) and sort them in descending order

    arr = freq.values()

    arr.sort()

    arr.reverse()

    # 26 * (count of most common letter) + (25 * next most common) + ...

    values_and_counts = zip(range(26, 0, -1), arr)

    return sum(value * count for value, count in values_and_counts)

def isBalanced(message):

    minOpen = 0

    maxOpen = 0

    for i in xrange(len(message)):

        if message[i] == '(':

            maxOpen += 1

            if i != 0 and message[i-1] != ':':

                minOpen += 1

        elif message[i] == ')':

            minOpen = max(0, minOpen-1)

            if i != 0 and message[i-1] != ':':

                maxOpen -= 1

                if maxOpen < 0:

                    break

    if maxOpen >= 0 and minOpen == 0:

        return "YES"

    else:

        return "NO"

//这里用到了set和multiset来维护数集
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <queue>

using namespace std;

int M[200020];

int main() {
  int N; cin >> N;
  for(int t = 0; t < N; t++) {
    int n, k; cin >> n >> k; n--;
    int a, b, c, r; cin >> a >> b >> c >> r;

    M[0] = a;
    for(int i = 1; i < k; i++) {
      M[i] = (1ll * b * M[i - 1] + c) % r;
    }

    set<int> st;
    for(int i = 0; i <= k; i++) st.insert(i);
    for(int i = 0; i < k; i++) st.erase(M[i]);

    multiset<int> dupst;
    for(int i = 0; i < k; i++) dupst.insert(M[i]);

    for(int i = k; i <= 2 * k; i++) {
      M[i] = *st.begin();
      st.erase(st.begin());

      if(i < 2 * k) {
        dupst.erase(dupst.find(M[i - k]));
        if(M[i - k] <= k && dupst.find(M[i - k]) == dupst.end()) {
          st.insert(M[i - k]);
        }
      }
    }

    cout << "Case #" << (t + 1) << ": ";
    if(n <= 2 * k) {
      cout << M[n] << endl;
    } else {
      cout << M[k + (n - 2 * k - 1) % (k + 1)] << endl;
    }
  }
  return 0;
}