PAT 2019 春

　　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。

 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <vector>

 using namespace std;
 bool isPrime(int n)
 {
     if(n < )
         return false;
     int tmpNum = (int)sqrt(1.0*n);
     for(int i = ; i <= tmpNum; ++ i)
     {
         if(n % i == )
             return false;
     }
     return true;
 }
 int main()
 {
     int tmpNum, prime1 = -;
     cin >> tmpNum;
     if(isPrime(tmpNum))
     {
         if(tmpNum >  && isPrime(tmpNum-))
             prime1 = tmpNum-;
         else if(isPrime(tmpNum+))
             prime1 = tmpNum + ;
     }
     prime1 == - ? cout << "No" << endl : cout << "Yes" << endl;
     for(int i = tmpNum-; prime1 == -;i ++)
     {
         if(isPrime(i) && isPrime(i+))
         {
             if(i > tmpNum)
                 prime1 = i;
             else
                 prime1 = i +;
         }

     }
     cout << prime1;
     return ;
 }

7-2 Anniversary

　　也是读懂题意就行。

 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <unordered_map>
 #include <string>
 #include <cstdlib>
 using namespace std;
 unordered_map<string, int> alumFlagMap;
 int main()
 {
     int n, tmpNum, oldGuestNum = , oldAlumNum = ;
     string tmpStr, oldGuest, oldAlum;
     cin >> n;
     while(n--)
     {
         cin >> tmpStr;
         alumFlagMap[tmpStr] = ;
     }
     cin >> n;
     int cnt = ;
     while(n--)
     {
         cin >> tmpStr;
         if(alumFlagMap[tmpStr] > )
         {
             tmpNum = atoi(tmpStr.substr(,).c_str());
             if(tmpNum < oldAlumNum)
             {
                 oldAlumNum = tmpNum;
                 oldAlum = tmpStr;
             }
             cnt ++;
         }
         else
         {
             tmpNum = atoi(tmpStr.substr(,).c_str());
             if(tmpNum < oldGuestNum)
             {
                 oldGuestNum = tmpNum;
                 oldGuest = tmpStr;
             }
         }
     }
     cout << cnt << endl;
     cnt ==  ? cout << oldGuest << endl : cout << oldAlum << endl;

     return ;
 }

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。

 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <unordered_map>
 #include <string>
 #include <cstdlib>
 #include <cstring>
 using namespace std;

 #define MAX_AMOUNT 1010
 const int INF = 0x7f7f7f7f;
 #define CLR(a,b) memset(a,b,sizeof(a))
 int callInfo[MAX_AMOUNT][MAX_AMOUNT];
 int fatherId[MAX_AMOUNT+];
 int K, N, M;
 vector<int> susPeopleVec;
 vector<int> phoneNumCnt[MAX_AMOUNT];
 unordered_map<int, int> susPeopleMap;
 int getFather(int u)
 {
     int tmpNum = u;
     while(fatherId[tmpNum] != tmpNum)
         tmpNum = fatherId[tmpNum];
     int tmpFa = tmpNum;
     tmpNum = u;
     while(fatherId[tmpNum] != tmpNum)
     {
         u = tmpNum;
         tmpNum = fatherId[tmpNum];
         fatherId[u] = tmpFa;
     }
     return tmpFa;
 }
 int main()
 {
     int tmpNum, tmpSt, tmpEnd, tmpTime, susCnt = ;
     cin >> K >> N >> M;
     CLR(callInfo, );
     while(M--)
     {
         cin >> tmpSt >> tmpEnd >> tmpTime;
         callInfo[tmpSt][tmpEnd] += tmpTime;
     }
     for(int i = ; i <= N; ++ i)
     {
         int shortCallCnt = ;
         int rebackCnt = ;
         for(int j = ; j <= N; ++ j)
         {
             if(callInfo[i][j] >  && callInfo[i][j] <= )
             {
                 shortCallCnt ++;
                 if(callInfo[j][i] > )
                     rebackCnt ++;
             }
         }
         if(shortCallCnt > K && 1.0*rebackCnt/shortCallCnt <= 0.2)
         {
             susPeopleMap[i] = ;susCnt++;susPeopleVec.push_back(i);
         }
     }
     for(int i = ; i <= N; ++ i)
     {
         fatherId[i] = i;
     }
     int index;
     for(int i = ; i < susPeopleVec.size(); ++ i)
     {
         index = susPeopleVec[i];
         for(int j = ; j <= N; ++ j)
         {
             if(susPeopleMap[j] == )
                 continue;
             if(callInfo[index][j] >  && callInfo[j][index] > )
             {
                 int tmpF1 = getFather(index);
                 int tmpF2 = getFather(j);
                 if(tmpF1 > tmpF2)
                 {
                     fatherId[tmpF1] = tmpF2;
                     getFather(index);
                 }
                 else
                 {
                     fatherId[tmpF2] = tmpF1;
                     getFather(j);
                 }
             }
         }
     }
     if(susPeopleVec.size() == )
     {
         cout << "None" << endl;
         return ;
     }
     for(int i = ; i < susPeopleVec.size(); ++ i)
         phoneNumCnt[fatherId[susPeopleVec[i]]].push_back(susPeopleVec[i]);
     for(int i = ; i <= N; ++ i)
     {
         if(phoneNumCnt[i].size() > )
         {
             bool symbolFlag = false;
             for(auto it = phoneNumCnt[i].begin(); it != phoneNumCnt[i].end(); ++ it)
             {
                 symbolFlag ? printf(" ") : symbolFlag = true;
                 printf("%d", *it);
             }
             printf("\n");
         }
     }
     return ;
 }

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）

 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <unordered_map>
 #include <string>
 #include <cstdlib>
 #include <cstring>
 using namespace std;

 #define MAX_AMOUNT 1010
 const int INF = 0x7f7f7f7f;
 #define CLR(a,b) memset(a,b,sizeof(a))
 int N, M;
 bool fullFlag = true;
 vector<int> postVec, inorderVec;
 vector<int> treeVec();
 unordered_map<int, int> nodeIndexMap;
 bool createTree(int inL, int inR, int postL, int postR, int nodeIndex)
 {
     if(inL > inR || postL > postR)
         return false;
     int tmpNum = postVec[postR];
     int index = inL;
     while(inorderVec[index] != tmpNum)
         index++;
     treeVec[nodeIndex] = tmpNum;
     nodeIndexMap[tmpNum] = nodeIndex;
     int childCnt = ;
     if(createTree(inL, index-, postL, postL+index--inL, nodeIndex*))
         childCnt ++;
     if(createTree(index+, inR, postR-inR+index, postR-, nodeIndex*+))
         childCnt ++;
     if(childCnt == )
         fullFlag = false;
     return true;
 }
 int getNumber(string tmpStr)
 {
     bool flag = false;
     int index = ;
     while(index < tmpStr.size())
     {
         if(isdigit(tmpStr[index]))
             break;
         index ++;
     }
     int endIndex = index;
     while(endIndex < tmpStr.size())
     {
         if(!isdigit(tmpStr[endIndex]))
         {
             break;
         }
         endIndex++;
     }
     return atoi(tmpStr.substr(index, endIndex-index).c_str());
 }
 int main()
 {
     cin >> N;
     int tmpNum;
     postVec.resize(N);
     inorderVec.resize(N);
     for(int i = ; i < N; ++ i)
         cin >> postVec[i];
     for(int i = ; i < N; ++ i)
         cin >> inorderVec[i];

     createTree(, N-, , N-, );
     cin >> M;
     getchar();
     string tmpStr;
     while(M--)
     {
         getline(cin, tmpStr);
         if(tmpStr[tmpStr.size()-] == 't')
         {
             getNumber(tmpStr) == treeVec[] ? printf("Yes\n") : printf("No\n");
         }
         else if(tmpStr[] == 'I')
         {
             fullFlag ? printf("Yes\n") : printf("No\n");
         }
         else if(tmpStr[tmpStr.size()-] == 's')
         {
             int tmpNum1 = getNumber(tmpStr.substr(,));
             int tmpNum2 = getNumber(tmpStr.substr(,));
             tmpNum1 = nodeIndexMap[tmpNum1];
             tmpNum2 = nodeIndexMap[tmpNum2];
             if((tmpNum1%== && tmpNum2-tmpNum1==)||tmpNum1%== && tmpNum1-tmpNum2==)
             {
                 printf("Yes\n");
             }
             else
                 printf("No\n");
         }
         else if(tmpStr[tmpStr.size()-] == 'l')
         {
             int tmpNum1 = getNumber(tmpStr.substr(,));
             int tmpNum2 = getNumber(tmpStr.substr(,));
             tmpNum1 = nodeIndexMap[tmpNum1];
             tmpNum2 = nodeIndexMap[tmpNum2];
             int cnt = ;
             while(tmpNum1 > )
             {
                 tmpNum1 /= ;
                 cnt++;
             }
             while(tmpNum2 > )
             {
                 tmpNum2 /= ;
                 cnt --;
             }
             if(cnt == )
                 printf("Yes\n");
             else
                 printf("No\n");
         }
         else
         {
             int tmpNum1 = getNumber(tmpStr.substr(,));
             int tmpNum2 = getNumber(tmpStr.substr(tmpStr.size()-,));
             tmpNum1 = nodeIndexMap[tmpNum1];
             tmpNum2 = nodeIndexMap[tmpNum2];
             if(tmpStr.find("left")!=tmpStr.npos)
             {
                 tmpNum1 == tmpNum2* ? printf("Yes\n") :printf("No\n");
             }
             else if(tmpStr.find("right")!=tmpStr.npos)
             {
                 tmpNum1 == tmpNum2*+ ? printf("Yes\n") :printf("No\n");
             }
             else
             {
                 tmpNum1 == tmpNum2/ ? printf("Yes\n") :printf("No\n");
             }
         }
     }
     return ;
 }