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10.Common Permutation

Time Limit: 4 sec

Description

Given two strings of lowercase letters, a and b, print the longest string x of lowercase letters such that there is a permutation of x that is a subsequence of a and there is a permutation of x that is a subsequence of b.

Input

Input file contains several lines of input. Consecutive two lines make a set of input. That means in the input file line 1 and 2 is a set of input, line 3 and 4 is a set of input and so on. The first line of a pair contains a and the second contains b. Each string is on a separate line and consists of at most 1000 lowercase letters.

Output

For each set of input, output a line containing x. If several x satisfy the criteria above, choose the first one in alphabetical order.

題意:輸入兩字串a、b，把兩字串擁有相同的單字印出來，並從a到z排列，若沒有該字元則跳過

1.統計兩個字串所有a至z字元個數

2.按照a至z的順序，印出字元(若count_a[i]=0就不會印出任何字元)

3.處理玩一組測資則換行

e↵\r\n
nw↵\r\n
et↵\r\n

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(){
  char a[1001],b[1001];
  int count_a[123],count_b[123];
  int i,j;
  while(gets(a)){
    memset(count_a,0,sizeof(count_a));
    memset(count_b,0,sizeof(count_b));
    for(i=0;i<strlen(a);i++)   //注意，for迴圈若沒用大括號，也不需要';'結尾
      count_a[a[i]]++;
    gets(b);
    for(i=0;i<strlen(b);i++)
      count_b[b[i]]++;
    for(i=97;i<123;i++){
      j=0;                                  //要在這個回圈內初始化
      while(j<count_a[i] && j<count_b[i]){  //j從0開始，如果count_a[i]和count_b[i]有數字就會印出來，如果沒有數字(=0)就會跳過
        printf("%c",i);
        j++;
      }
    }
      printf("\n");
  }
  return 0;
}
    

"R↵\r\n
Ie↵\r\n
 n↵\r\n
te↵\r\n
h ↵\r\n
iD↵\r\n
ne↵\r\n
kc↵\r\n
,a↵\r\n
 r↵\r\n
tt↵\r\n
he↵\r\n
es↵\r\n
r ↵\r\n
eo↵\r\n
fn↵\r\n
oc↵\r\n
re↵\r\n
e ↵\r\n
 s↵\r\n
Ia↵\r\n
 i↵\r\n
ad↵\r\n
m,↵\r\n
.↵\r\n
"↵\r\n

#include <iostream>
#include <cstring>
using namespace std;
int main(){
  char str[100][101];       //最多100行，每行最多100個字元
  int len[100],n=0,max=0;
  for(int i=0;i<100;i++){
    for(int j=0;j<101;j++){
      str[i][j]=0;
      len[n]=0;
    }
  }
  while(cin.getline(str[n],101)){
    len[n]=strlen(str[n]);
    if(len[n]>max)
      max = len[n];
    for(int add=len[n];add<max;add++){
      str[n][add]=' ';
      len[n]++;
    }
    n++;
  }
  for(int j=0;j<max;j++){
    for(int i=n-1;i>=0;i--)
      if(j<len[i])
        cout<<str[i][j];
    cout<<endl;
    
  }
  return 0;
}

12.TeX Quotes

題意:將該字串，第一個"改為``，第二個"改為''

``To be or not to be,'' quoth the Bard, ``that↵\r\n
is the question''.↵\r\n
The programming contestant replied: ``I must disagree.↵\r\n
To `C' or not to `C', that is The Question!''↵\r\n

13.Doom's Day Algorithm

Time Limit: 5 sec

Background

No. Doom's day algorithm is not a method to compute which day the world will end. It is an algorithm created by the mathematician John Horton Conway, to calculate which day of the week (Monday, Tuesday, etc.) corresponds to a certain date.

This algorithm is based in the idea of the doomsday, a certain day of the week which always occurs in the same dates. For example, 4/4 (the 4th of April), 6/6 (the 6th of June), 8/8 (the 8th of August), 10/10 (the 10th of October) and 12/12 (the 12th of December) are dates which always occur in doomsday. All years have their own doomsday.

In year 2011, doomsday is Monday. So all of 4/4, 6/6, 8/8, 10/10 and 12/12 are Mondays. Using that information, you can easily compute any other date. For example, the 13th of December 2011 will be Tuesday, the 14th of December 2011 will be Wednesday, etc.

Description

Other days which occur on doomsday are 5/9, 9/5, 7/11 and 11/7. Also, in leap years, we have the following doomsdays: 1/11 (the 11th of January) and 2/22 (the 22nd of Febrary), and in non-leap years 1/10 and 2/21.

Given a date of year 2011, you have to compute which day of the week it occurs.

Input

The input can contain different test cases. The first line of the input indicates the number of test cases.

For each test case, there is a line with two numbers: M D. M represents the month (from 1 to 12) and D represents the day (from 1 to 31). The date will always be valid.

Output

For each test case, you have to output the day of the week where that date occurs in 2011. The days of the week will be: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.

題意:給日期算出是星期幾

1.根據題目測資推算，2011/01/06是星期四，所以2010/12/31是星期五，並以此為起點

2.(迴圈算出月份總天數，再加上該月已過的天數)，除以7，若餘數為0，則該日為星期日

#include <iostream>
using namespace std;
int main(){
  char week[7][10]={"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};  //從星期日開始
  int n;
  int Month[]={31,28,31,30,31,30,31,31,30,31,30,31};
  cin>>n;
  while(n--){
    int k=5;                     //起始值(由1/1日星期幾校正)
    int month,day;
    cin>>month>>day;
    for(int i=1;i<month;i++)
      k=k+Month[i-1];            //month個月的天數加總，再加上起始值
    k=(k+day)%7;                 //month個月的天數加總+起始值+此次輸入的day值)除以7取得餘數
    cout<<week[k]<<endl;
  }
  return 0;
}
     

#include <iostream>
#include <set>
using namespace std;
int main(){
  int n;
  while(cin>>n){
    set<int> tank;
    int a;cin>>a;
    for(int i=1;i<n;i++){
      int b;cin>>b;
      int d=(b-a<0?a-b:b-a);
      
      if(d && d<n)tank.insert(d);
      a=b;
    }
    if(tank.size()==n-1)cout<<"Jolly";
    else  cout<<"Not jolly";
    cout << endl;
  }
  return 0;
}
    

#include <iostream>
using namespace std;
int main(){
  int d,a,b,c,s; //d有幾筆資料,a、b為兩數字，c為項數，s為數列合
  int ans=0;     //幫case計數
  cin>>d;
  while(d--){
  while(cin>>a>>b){
    if(a%2==0)
      a++;
    if(b%2==0)
      b--;
    c = ((b-a)/2)+1;
    s =c*(a+b)/2;
    ans++;
    cout<<"Case "<<ans<<": "<<s<<endl;
    
    }
    
  }
  return 0;
}