PAT 2019 春

Posted 2022-07-19 codewars

　　算是第二次做这套题吧，感觉从上次考试到现在自己有了挺大提高，提前30min做完了。

7-1 Sexy Primes

　　读懂题意就行。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cmath>
 4 #include <algorithm>
 5 #include <vector>
 6 
 7 using namespace std;
 8 bool isPrime(int n)
 9 
10     if(n < 2)
11         return false;
12     int tmpNum = (int)sqrt(1.0*n);
13     for(int i = 2; i <= tmpNum; ++ i)
14     
15         if(n % i == 0)
16             return false;
17     
18     return true;
19 
20 int main()
21 
22     int tmpNum, prime1 = -1;
23     cin >> tmpNum;
24     if(isPrime(tmpNum))
25     
26         if(tmpNum > 7 && isPrime(tmpNum-6))
27             prime1 = tmpNum-6;
28         else if(isPrime(tmpNum+6))
29             prime1 = tmpNum + 6;
30     
31     prime1 == -1 ? cout << "No" << endl : cout << "Yes" << endl;
32     for(int i = tmpNum-5; prime1 == -1;i ++)
33     
34         if(isPrime(i) && isPrime(i+6))
35         
36             if(i > tmpNum)
37                 prime1 = i;
38             else
39                 prime1 = i +6;
40         
41             
42     
43     cout << prime1;
44     return 0;
45 

View Code

7-2 Anniversary

　　也是读懂题意就行。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cmath>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <unordered_map>
 7 #include <string>
 8 #include <cstdlib>
 9 using namespace std;
10 unordered_map<string, int> alumFlagMap;
11 int main()
12 
13     int n, tmpNum, oldGuestNum = 999999999, oldAlumNum = 999999999;
14     string tmpStr, oldGuest, oldAlum;
15     cin >> n;
16     while(n--)
17     
18         cin >> tmpStr;
19         alumFlagMap[tmpStr] = 1;
20     
21     cin >> n;
22     int cnt = 0;
23     while(n--)
24     
25         cin >> tmpStr;
26         if(alumFlagMap[tmpStr] > 0)
27         
28             tmpNum = atoi(tmpStr.substr(6,8).c_str());
29             if(tmpNum < oldAlumNum)
30             
31                 oldAlumNum = tmpNum;
32                 oldAlum = tmpStr;
33             
34             cnt ++;
35         
36         else
37         
38             tmpNum = atoi(tmpStr.substr(6,8).c_str());
39             if(tmpNum < oldGuestNum)
40             
41                 oldGuestNum = tmpNum;
42                 oldGuest = tmpStr;
43             
44         
45     
46     cout << cnt << endl;
47     cnt == 0 ? cout << oldGuest << endl : cout << oldAlum << endl;
48 
49     return 0;
50 

View Code

7-3 Telefraud Detection

　　这道题目理清条件就行，先找到嫌疑人，在根据嫌疑人找团伙，团伙进行并查集处理。

　　   　1.嫌疑人：给大于K个人打短电话，且不超过20%的人回拨。

　　    2.团伙：两个嫌疑人互相打过电话。

　　　3.短电话：A拨打B电话的总时长不超过5分钟，认为A向B打短电话。

 

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <cmath>
  4 #include <algorithm>
  5 #include <vector>
  6 #include <unordered_map>
  7 #include <string>
  8 #include <cstdlib>
  9 #include <cstring>
 10 using namespace std;
 11 
 12 #define MAX_AMOUNT 1010
 13 const int INF = 0x7f7f7f7f;
 14 #define CLR(a,b) memset(a,b,sizeof(a))
 15 int callInfo[MAX_AMOUNT][MAX_AMOUNT];
 16 int fatherId[MAX_AMOUNT+1];
 17 int K, N, M;
 18 vector<int> susPeopleVec;
 19 vector<int> phoneNumCnt[MAX_AMOUNT];
 20 unordered_map<int, int> susPeopleMap;
 21 int getFather(int u)
 22 
 23     int tmpNum = u;
 24     while(fatherId[tmpNum] != tmpNum)
 25         tmpNum = fatherId[tmpNum];
 26     int tmpFa = tmpNum;
 27     tmpNum = u;
 28     while(fatherId[tmpNum] != tmpNum)
 29     
 30         u = tmpNum;
 31         tmpNum = fatherId[tmpNum];
 32         fatherId[u] = tmpFa;
 33     
 34     return tmpFa;
 35 
 36 int main()
 37 
 38     int tmpNum, tmpSt, tmpEnd, tmpTime, susCnt = 0;
 39     cin >> K >> N >> M;
 40     CLR(callInfo, 0);
 41     while(M--)
 42     
 43         cin >> tmpSt >> tmpEnd >> tmpTime;
 44         callInfo[tmpSt][tmpEnd] += tmpTime;
 45     
 46     for(int i = 1; i <= N; ++ i)
 47     
 48         int shortCallCnt = 0;
 49         int rebackCnt = 0;
 50         for(int j = 1; j <= N; ++ j)
 51         
 52             if(callInfo[i][j] > 0 && callInfo[i][j] <= 5)
 53             
 54                 shortCallCnt ++;
 55                 if(callInfo[j][i] > 0)
 56                     rebackCnt ++;
 57             
 58         
 59         if(shortCallCnt > K && 1.0*rebackCnt/shortCallCnt <= 0.2)
 60         
 61             susPeopleMap[i] = 1;susCnt++;susPeopleVec.push_back(i);
 62         
 63     
 64     for(int i = 1; i <= N; ++ i)
 65     
 66         fatherId[i] = i;
 67     
 68     int index;
 69     for(int i = 0; i < susPeopleVec.size(); ++ i)
 70     
 71         index = susPeopleVec[i];
 72         for(int j = 1; j <= N; ++ j)
 73         
 74             if(susPeopleMap[j] == 0)
 75                 continue;
 76             if(callInfo[index][j] > 0 && callInfo[j][index] > 0)
 77             
 78                 int tmpF1 = getFather(index);
 79                 int tmpF2 = getFather(j);
 80                 if(tmpF1 > tmpF2)
 81                 
 82                     fatherId[tmpF1] = tmpF2;
 83                     getFather(index);
 84                 
 85                 else
 86                 
 87                     fatherId[tmpF2] = tmpF1;
 88                     getFather(j);
 89                 
 90             
 91         
 92     
 93     if(susPeopleVec.size() == 0)
 94     
 95         cout << "None" << endl;
 96         return 0;
 97     
 98     for(int i = 0; i < susPeopleVec.size(); ++ i)
 99         phoneNumCnt[fatherId[susPeopleVec[i]]].push_back(susPeopleVec[i]);
100     for(int i = 1; i <= N; ++ i)
101     
102         if(phoneNumCnt[i].size() > 0)
103         
104             bool symbolFlag = false;
105             for(auto it = phoneNumCnt[i].begin(); it != phoneNumCnt[i].end(); ++ it)
106             
107                 symbolFlag ? printf(" ") : symbolFlag = true;
108                 printf("%d", *it);
109             
110             printf("\n");
111         
112     
113     return 0;
114 

View Code

 

7-4 Structure of a Binary Tree

　　这种判断是否正确的题目，一般选择比较适合查找的顺序存储结构。

　　首先根据题目给出的后序和先序顺序，在数组中将该二叉树建立出来。

　　之后便是提取题目给出的结论中的数字，这个自己写个函数就可以搞定。

　　最后便是利用string.find()的查找功能确定结论是哪一种，然后进行判断（因为是顺序存储，判断会相当简单）。

　　（代码有点啰嗦，赶时间哈）

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <cmath>
  4 #include <algorithm>
  5 #include <vector>
  6 #include <unordered_map>
  7 #include <string>
  8 #include <cstdlib>
  9 #include <cstring>
 10 using namespace std;
 11 
 12 #define MAX_AMOUNT 1010
 13 const int INF = 0x7f7f7f7f;
 14 #define CLR(a,b) memset(a,b,sizeof(a))
 15 int N, M;
 16 bool fullFlag = true;
 17 vector<int> postVec, inorderVec;
 18 vector<int> treeVec(10000);
 19 unordered_map<int, int> nodeIndexMap;
 20 bool createTree(int inL, int inR, int postL, int postR, int nodeIndex)
 21 
 22     if(inL > inR || postL > postR)
 23         return false;
 24     int tmpNum = postVec[postR];
 25     int index = inL;
 26     while(inorderVec[index] != tmpNum)
 27         index++;
 28     treeVec[nodeIndex] = tmpNum;
 29     nodeIndexMap[tmpNum] = nodeIndex;
 30     int childCnt = 0;
 31     if(createTree(inL, index-1, postL, postL+index-1-inL, nodeIndex*2))
 32         childCnt ++;
 33     if(createTree(index+1, inR, postR-inR+index, postR-1, nodeIndex*2+1))
 34         childCnt ++;
 35     if(childCnt == 1)
 36         fullFlag = false;
 37     return true;
 38 
 39 int getNumber(string tmpStr)
 40 
 41     bool flag = false;
 42     int index = 0;
 43     while(index < tmpStr.size())
 44     
 45         if(isdigit(tmpStr[index]))
 46             break;
 47         index ++;
 48     
 49     int endIndex = index;
 50     while(endIndex < tmpStr.size())
 51     
 52         if(!isdigit(tmpStr[endIndex]))
 53         
 54             break;
 55         
 56         endIndex++;
 57     
 58     return atoi(tmpStr.substr(index, endIndex-index).c_str());
 59 
 60 int main()
 61 
 62     cin >> N;
 63     int tmpNum;
 64     postVec.resize(N);
 65     inorderVec.resize(N);
 66     for(int i = 0; i < N; ++ i)
 67         cin >> postVec[i];
 68     for(int i = 0; i < N; ++ i)
 69         cin >> inorderVec[i];
 70 
 71     createTree(0, N-1, 0, N-1, 1);
 72     cin >> M;
 73     getchar();
 74     string tmpStr;
 75     while(M--)
 76     
 77         getline(cin, tmpStr);
 78         if(tmpStr[tmpStr.size()-1] == ‘t‘)
 79         
 80             getNumber(tmpStr) == treeVec[1] ? printf("Yes\n") : printf("No\n");
 81         
 82         else if(tmpStr[0] == ‘I‘)
 83         
 84             fullFlag ? printf("Yes\n") : printf("No\n");
 85         
 86         else if(tmpStr[tmpStr.size()-1] == ‘s‘)
 87         
 88             int tmpNum1 = getNumber(tmpStr.substr(0,4));
 89             int tmpNum2 = getNumber(tmpStr.substr(4,10));
 90             tmpNum1 = nodeIndexMap[tmpNum1];
 91             tmpNum2 = nodeIndexMap[tmpNum2];
 92             if((tmpNum1%2==0 && tmpNum2-tmpNum1==1)||tmpNum1%2==1 && tmpNum1-tmpNum2==1)
 93             
 94                 printf("Yes\n");
 95             
 96             else
 97                 printf("No\n");
 98         
 99         else if(tmpStr[tmpStr.size()-1] == ‘l‘)
100         
101             int tmpNum1 = getNumber(tmpStr.substr(0,4));
102             int tmpNum2 = getNumber(tmpStr.substr(4,10));
103             tmpNum1 = nodeIndexMap[tmpNum1];
104             tmpNum2 = nodeIndexMap[tmpNum2];
105             int cnt = 0; 
106             while(tmpNum1 > 0)
107             
108                 tmpNum1 /= 2;
109                 cnt++;
110             
111             while(tmpNum2 > 0)
112             
113                 tmpNum2 /= 2;
114                 cnt --;
115             
116             if(cnt == 0)
117                 printf("Yes\n");
118             else
119                 printf("No\n");
120         
121         else
122         
123             int tmpNum1 = getNumber(tmpStr.substr(0,5));
124             int tmpNum2 = getNumber(tmpStr.substr(tmpStr.size()-10,10));
125             tmpNum1 = nodeIndexMap[tmpNum1];
126             tmpNum2 = nodeIndexMap[tmpNum2];
127             if(tmpStr.find("left")!=tmpStr.npos)
128             
129                 tmpNum1 == tmpNum2*2 ? printf("Yes\n") :printf("No\n");
130             
131             else if(tmpStr.find("right")!=tmpStr.npos)
132             
133                 tmpNum1 == tmpNum2*2+1 ? printf("Yes\n") :printf("No\n");
134             
135             else
136             
137                 tmpNum1 == tmpNum2/2 ? printf("Yes\n") :printf("No\n");
138             
139         
140     
141     return 0;
142 

View Code