2021暑期积分赛一

A - Constructing the Array （CF353D）
B - Taxi （CF158B）
C - Running Median HDU - 3282
D - What Are You Talking About HDU - 1075
E - A + B Problem II HDU - 1002
F - 首字母变大写 HDU - 2026
G - Solving Order HDU - 5702
H - 排序 HDU - 1106
I - 前m大的数 HDU - 1280

A - Constructing the Array （CF353D）

There is an array a of size n consiting of all zeros. You have to perform n operations on this array. On the ith operation, you need to perform the following steps:

Choose the maximum size subarray consisting of all zeros. If there are multiple such subarrays of the same size, select the leftmost one.
Let the subarray be [l,r]. If the length of the subarray is and odd number, then perform
a[(l + r) / 2] = i [that is assign the value i to the ((l + r) / 2)th index of the array], otherwise perform a[(l + r - 1) / 2] = i [that is assign the value i to the ((l + r - 1) / 2)th index of the array]
For example, let the initial array a be of length 5 (a = [0, 0, 0, 0, 0]). Then the operations are

The chosen subarray is [1, 5] and we assign a[3] = 1. The array becomes [0, 0, 1, 0, 0]
The chosen subarray is [1, 2] and we assign a[1] = 2. The array becomes [2, 0, 1, 0, 0]
The chosen subarray is [4, 5] and we assign a[4] = 3. The array becomes [2, 0, 1, 3, 0]
The chosen subarray is [2, 2] and we assign a[2] = 4. The array becomes [2, 4, 1, 3, 0]
The chosen subarray is [5, 5] and we assign a[5] = 5. The array becomes [2, 4, 1, 3, 5]
Your task is to find the final array after all n operations. It is guaranteed that the answer always exists and is unique.

Input
The first line of the input contains an integer t (1 ≤ t ≤ 104) — the number of test cases. Then t test cases follow.

Each test case contains only one line containing one integer n (1 ≤ n ≤ 2 × 105) — the length of the array a.

It is guaranteed that the sum of n over all test cases does not exceed 2 × 105 (∑n ≤ 2 × 105)

Output
For each test case, print the final array after performing all the n operations.

Sample Input
6
1
2
3
4
5
6
Sample Output
1
1 2
2 1 3
3 1 2 4
2 4 1 3 5
3 4 1 5 2 6
AC代码

#include <iostream>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
struct node
{int l,r;bool operator<(const node&o)const{if(r-l==o.r-o.l)    return l>o.r;else    return r-l<o.r-o.l;}
};
priority_queue<node> dq;
int ans[200010];
int main()
{int t;cin>>t;while(t--){int n;cin>>n;memset(ans,0,sizeof(ans));dq.push({1,n});int k=0;while(!dq.empty()){node q=dq.top();dq.pop();int l=q.l;int r=q.r;int m=(l+r)/2;k++;ans[m]=k;if(l!=r){if(m==l) dq.push({m+1,r});else{dq.push({l,m-1});dq.push({m+1,r});}}}cout<<ans[1];for(int i=2;i<=n;i++)    cout<<" "<<ans[i];cout<<endl;}return 0;
}

B - Taxi （CF158B）

学校有n个小组外出游玩，第i个小组有si (1 ≤ si ≤ 4)个人，同学们想要打出租出回来，每辆车最多坐4个人，每个小组必须坐在同一辆车上，求最少需要多少辆车才能让同学们都回来；
Input
第一行一个整数n (1 ≤ n ≤ 1e5)，接下来n个数表示si (1 ≤ si ≤ 4)，
Output
输出一个整数表示结果

Input
5
1 2 4 3 3
Output
4
AC代码

#include <iostream>
using namespace std;
int main()
{int n;cin>>n;int cnt=0;int one=0,two=0,three=0,four=0;while(n--){int s;cin>>s;switch(s){case 1:one++;break;case 2:two++;break;case 3:three++;break;case 4:four++;break;}}cnt=four+three+two/2; //最少这么多，再把多余的1拼入three和two if(three<one)   one-=three;//1在拼入three后仍有剩余 else   one=0;if(two&1)    //two是奇数，剩余一组two未算上 {cnt++;if(one-2>0)  one-=2;    //如果1的个数大于2，剩余1的个数为one-2else one=0; }cnt+=one/4;      //最后剩余的1互相组队 if(one%4!=0)  cnt++;cout<<cnt<<endl;return 0;
}

C - Running Median HDU - 3282

For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far.
Input
The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by an odd decimal integer M, (1 ≤ M ≤ 9999), giving the total number of signed integers to be processed.
The remaining line(s) in the dataset consists of the values, 10 per line, separated by a single space.
The last line in the dataset may contain less than 10 values.
Output
For each data set the first line of output contains the data set number, a single space and the number of medians output (which should be one-half the number of input values plus one). The output medians will be on the following lines, 10 per line separated by a single space. The last line may have less than 10 elements, but at least 1 element. There should be no blank lines in the output.
Sample Input
3
1 9
1 2 3 4 5 6 7 8 9
2 9
9 8 7 6 5 4 3 2 1
3 23
23 41 13 22 -3 24 -31 -11 -8 -7
3 5 103 211 -311 -45 -67 -73 -81 -99
-33 24 56
Sample Output
1 5
1 2 3 4 5
2 5
9 8 7 6 5
3 12
23 23 22 22 13 3 5 5 3 -3
-7 -3
AC代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <map>
#include <set>
#include <queue>
#include <deque>
using namespace std;
int x;
int main()
{int p;cin>>p;while(p--){int id,n;cin>>id>>n;cout<<id<<' '<<(n+1)/2<<endl;//用两个优先队列存放数组 priority_queue<int,vector<int>,greater<int> > gq;    //升序，大先出 priority_queue<int,vector<int>,less<int> > lq;    //降序，小先出 int cnt=0; //记录输出数字的个数，用于判断是否输出endl for(int i=1;i<=n;i++){cin>>x;if(i==1)    lq.push(x); //第一个数直接丢进lq else{if(x>lq.top()) gq.push(x);  else   lq.push(x);}            if((int)gq.size()-(int)lq.size()>=1)    //如果gq中元素比lq中多 {                                        //把gq中top元素挪进lq int temp=gq.top();                 //始终维护lq中元素比gq多 lq.push(temp);gq.pop();}if((int)lq.size()-(int)gq.size()>1)      //如果lq中元素比gq多 {                                     //那么始终维护lq中元素之比gq中多一 int temp=lq.top();                    //多出来的top元素就是所求中位数 gq.push(temp);lq.pop();} if(i%2==1)    //如果i是奇数 {cout<<lq.top();cnt++;if(cnt%10==0)  cout<<endl;       //当输出十个数字 换行 else if(i==n)    cout<<endl;       //所有所求数字输出完毕，换行 else cout<<" ";              //否则，输出空格 }}}return 0;
}

D - What Are You Talking About HDU - 1075

小智有一只神奇的电耗子（就决定是你了，皮卡丘！），它每天都会写很多东西，但是他看不懂这只黄皮耗子整天在写些什么。有一天他突然收到一个快递。他拆开后发现是一个字典，他就好奇地翻了起来，结果发现把皮卡丘写的东西和字典里的东西一对比，竟然能翻译成英文，他很想把皮卡丘写的东西全部翻一下来，你能帮他把皮卡丘写的东西翻译成英文吗？
Input
本题只有一组数据，所有单词只包含小写字母。测试用例由两部分组成，字典部分以一个单行字符串“START”开头，此字符串应该被忽略。然后下面每行包含两个单词，第一个是英文单词，第二个是皮卡丘的语言。字典以一个单行字符串“END”结束。字典不包含“START”和“END”。皮卡丘的文章同样以“START”开头，以“END”结束，然后是一篇皮卡丘写的文章。你应该将文章翻译成英文。如果你在字典中找到该单词，则应将其翻译并将新单词写入你的翻译，如果你无法在字典中找到该单词，则无需翻译它，只需将旧单词复制到你的翻译中即可。空格，制表符和回车及所有标点符号不应该被翻译。每行最多包含3000个字符
Output
你应该输出翻译的文章
Sample Input
START
from fiwo
hello difh
dream riwosf
you fnnvk
like fiiwj
END
START
difh, i’m fiwo riwosf.
i fiiwj fnnvk!
END
Sample Output
hello, i’m from dream.
i like you!
AC代码

#include <iostream>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;
map<string,string> mp;
string s1,s2,s;
int judge(char c)
{if(c>='a'&&c<='z') return 1;else return 0;
}
int main()
{while(cin>>s1){if(s1=="START")   continue;if(s1=="END")  break;cin>>s2;mp[s2]=s1;}getchar(); //cin读不入回车，需要加个getchar读入回车，否则会直接getline一行空白 while(getline(cin,s)){if(s=="START")   continue;if(s=="END")   break;int i=0;while(i<s.size()){if(judge(s[i])){string str="";while(judge(s[i])){str=str+s[i];i++;}if(mp[str]!="")    cout<<mp[str];else    cout<<str;}else{cout<<s[i];i++;}  }cout<<endl;}return 0;
}

E - A + B Problem II HDU - 1002

Input
输入数据的第一行为一个整数T（1≤T≤20），表示测试数据总数，紧接着的T行数据，每行包含由空格隔开的两个整数a和b，每组数据占一行。注意了，a和b可能非常大，大到超过32位整数可以表示的范围，我们假定a和b的位数不超过1000。
Output
对于每组数据，你需要输出两行，第一行显示"Case #:"（注意输出为英文字符），第二行为一个等式"a + b = Sum"，这里的Sum就是指a + b的结果（注意这个等式中的空格）。请在每两组输出数据间输出一个空行。
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
AC代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <map>
#include <set>
#include <cmath>
using namespace std;
int t;
string a,b;
int aa[1111],bb[1111];
int cc[1111];
int k=0;
int main()
{cin>>t;while(t--){cin>>a>>b;int la=a.size(),lb=b.size();k++;int lm=max(la,lb);memset(aa,0,sizeof(aa));  //注意这里是将数组全部内存空间置0，不是简单的到la、lb memset(bb,0,sizeof(bb));  //因为如果上次的长度大，数据可能会对这次产生影响 for(int i=0;i<la;i++)    aa[i]=a[la-i-1]-'0'; //“1234”存放数字的数组第一位应该是字符串的最后一位4 for(int i=0;i<lb;i++)  bb[i]=b[lb-i-1]-'0';int jin=0,cnt=0;for(int i=0;i<lm;i++)    //注意这里是小于lm    因为前述步骤已经对多余位置置0，所以不会有影响 {                                         //“123”+“3” ==>  123+003 int sum=aa[i]+bb[i]+jin;jin=sum/10;cc[i]=sum%10;cnt++;}if(jin)       //如果最后一位进位不为0 {cc[cnt]=1;cnt++;} cout<<"Case "<<k<<':'<<endl;cout<<a<<" + "<<b<<" = ";for(int i=cnt-1;i>=0;i--)   cout<<cc[i];  //倒序输出 cout<<endl;if(t>0)  cout<<endl;}return 0;
}

F - 首字母变大写 HDU - 2026

输入一个英文句子，将每个单词的第一个字母改成大写字母。
Input
输入数据包含多个测试实例，每个测试实例是一个长度不超过100的英文句子，占一行。
Output
请输出按照要求改写后的英文句子。
Sample Input
i like acm
i want to get an accepted
Sample Output
I Like Acm
I Want To Get An Accepted
AC代码

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
string s;
int main()
{while(getline(cin,s)) //读入一行带空格字符串 {s[0]-=32;for(int i=1;i<s.size();i++)    //字符串长度用.size() {if(s[i-1]==' ')    s[i]-=32;} cout<<s<<endl;      }   return 0;
}
/*
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
string s;
int main()
{while(getline(cin,s)) {s[0]-=32;for(int i=1;i<s.size();i++){if(s[i-1]==' ') s[i]-=32;} cout<<s<<endl;      }   return 0;
}
*/

G - Solving Order HDU - 5702

Input
The first line is an integer T which indicates the case number.
And as for each case, the first line is an integer n, which is the number of problems.
Then there are n lines followed, with a string and an integer in each line, in the i-th line, the string means the color of ballon for the i-th problem, and the integer means the ballon numbers.

It is guaranteed that:
T is about 100.
1≤n≤10.
1≤ string length ≤10.
1≤ bolloon numbers ≤83.(there are 83 teams :p)
For any two problems, their corresponding colors are different.
For any two kinds of balloons, their numbers are different.
Output
For each case, you need to output a single line.
There should be n strings in the line representing the solving order you choose.
Please make sure that there is only a blank between every two strings, and there is no extra blank.
Sample Input
3
3
red 1
green 2
yellow 3
1
blue 83
2
red 2
white 1
Sample Output
yellow green red
blue
red white
AC代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <map>
#include <set>
using namespace std;
struct Bo
{string color;int num;
}ballon[11];
bool cmp(Bo a,Bo b)
{return a.num>b.num;
}
int main()
{int t;cin>>t;while(t--){int n;cin>>n;for(int i=0;i<n;i++)cin>>ballon[i].color>>ballon[i].num;sort(ballon,ballon+n,cmp);cout<<ballon[0].color;for(int i=1;i<n;i++)cout<<" "<<ballon[i].color;cout<<endl;}return 0;
}

H - 排序 HDU - 1106

输入一行数字，如果我们把这行数字中的‘5’都看成空格，那么就得到一行用空格分割的若干非负整数（可能有些整数以‘0’开头，这些头部的‘0’应该被忽略掉，除非这个整数就是由若干个‘0’组成的，这时这个整数就是0）。

你的任务是：对这些分割得到的整数，依从小到大的顺序排序输出。

Input
输入包含多组测试用例，每组输入数据只有一行数字（数字之间没有空格），这行数字的长度不大于1000。

输入数据保证：分割得到的非负整数不会大于100000000；输入数据不可能全由‘5’组成。
Output
对于每个测试用例，输出分割得到的整数排序的结果，相邻的两个整数之间用一个空格分开，每组输出占一行。
Sample Input
0051231232050775
Sample Output
0 77 12312320
AC代码

#include <iostream>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
vector<int> v;
string s;
int main()
{while(cin>>s){   v.clear();int l=s.size();int ind=0;   //下标 while(ind<l){if(s[ind]=='5')    ind++;    //如果该字符等于‘5’，ind++，遍历下一个字符 else{string x=""; //空字符串，用于存放分割出来的字符串 int cnt=0;while(s[ind]!='5'&&ind<l){x=x+s[ind];cnt++;ind++;}int sum=0;for(int i=0;i<cnt;i++) //把字符串转化为数字 {int xx=x[i]-'0';sum=sum*10+xx;}v.push_back(sum);}}sort(v.begin(),v.end());vector<int>::iterator it;it=v.begin();cout<<*it;for(it++;it!=v.end();it++)cout<<' '<<*it;cout<<endl;}return 0;
}

I - 前m大的数 HDU - 1280

给定一个包含N(N<=3000)个正整数的序列，每个数不超过5000，对它们两两相加得到的N*(N-1)/2个和，求出其中前M大的数(M<=1000)并按从大到小的顺序排列。
Input
输入可能包含多组数据，其中每组数据包括两行：
第一行两个数N和M，
第二行N个数，表示该序列。

Output
对于输入的每组数据，输出M个数，表示结果。输出应当按照从大到小的顺序排列。
Sample Input
4 4
1 2 3 4
4 5
5 3 6 4
Sample Output
7 6 5 5
11 10 9 9 8
AC代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cstdio>
using namespace std;
int n,m;
int a[3333];
int v[10000000];    //注意v的取值范围
bool cmp(int a,int b)
{return a>b;
}
int main()
{while(scanf("%d%d",&n,&m)!=EOF){for(int i=0;i<n;i++)  scanf("%d",&a[i]);int k=0;for(int i=0;i<n-1;i++){for(int j=i+1;j<n;j++){v[k]=a[i]+a[j];k++;}}sort(v,v+k,cmp);printf("%d",v[0]);for(int it=1;it<m;it++)printf(" %d",v[it]);printf("\n");    }return 0;
}