PAT A1005-1008

A 1005 Spell It Right (20 point(s))

　　25分的题目，比较简单，注意N的范围，用字符串处理即可。

 #include <iostream>

 #include <cstdio>

 #include <algorithm>

 #include <vector>

 #include <string>

 #include <cstring>

 using namespace std;

 int main()

 {

     const char * numStr[]={"zero","one","two","three","four","five","six","seven","eight","nine"};

     string tmpStr;

     cin >> tmpStr;

     int sum = ;

     for(int i = ; i < tmpStr.size(); ++ i)

         sum += tmpStr[i]-'';

     char charArray[];

     sprintf(charArray, "%d", sum);

     for(int i = ; i < strlen(charArray); ++ i)

     {

         if(i > ) printf(" ");

         printf("%s", numStr[charArray[i]-'']);

     }

     return ;

 }

A 1006 1006 Sign In and Sign Out (25 point(s))

 #include <iostream>

 #include <cstdio>

 #include <algorithm>

 #include <vector>

 #include <string>

 #include <cstring>

 using namespace std;

 int main()

 {

     int M, tmpHour, tmpMinute, tmpSecond, unLockTime=*, lockTime=-;

     cin >> M;

     string tmpStr, unLockId, lockId;

     for(int i = ; i < M; ++i)

     {

         cin >> tmpStr;

         scanf("%d:%d:%d", &tmpHour, &tmpMinute, &tmpSecond);

         tmpHour = tmpHour*+tmpMinute*+tmpSecond;

         if(tmpHour < unLockTime)

         {

             unLockId = tmpStr;

             unLockTime = tmpHour;

         }

         scanf("%d:%d:%d", &tmpHour, &tmpMinute, &tmpSecond);

         tmpHour = tmpHour*+tmpMinute*+tmpSecond;

         if(tmpHour > lockTime)

         {

             lockId = tmpStr;

             lockTime = tmpHour;

         }

     }

     cout << unLockId << " " << lockId;

     return ;

 }

A 1007 Maximum Subsequence Sum (25 point(s))

　　经典的最大子序列和问题，注意 全为负数 和 只有负数和0 这两种情况即可，理不清题意可能会有一两个点过不去。

 #include <iostream>

 #include <cstdio>

 #include <algorithm>

 #include <vector>

 #include <string>

 #include <cstring>

 using namespace std;

 int main()

 {

     int K, tmpSt, resSt, resEnd, tmpNum, sum = -, maxSum = -;

     cin >> K;

     bool negFlag = true;

     for(int i = ; i < K; ++ i)

     {

         cin >> tmpNum;

         if(i==)

             resSt = tmpNum;

         if(tmpNum >= )

             negFlag = false;

         if(sum < )

         {

             sum = ;

             tmpSt = tmpNum;

         }

         sum += tmpNum;

         if(sum > maxSum)

         {

             maxSum = sum;

             resSt = tmpSt;

             resEnd = tmpNum;

         }

     }

     if(negFlag)

         cout <<  << " " << resSt << " " << tmpNum;

     else

         cout << maxSum << " " << resSt << " " << resEnd;

     return ;

 }

A 1008 Elevator (20 point(s))

　　一个直观的数学计算问题

 #include <iostream>

 #include <cstdio>

 #include <algorithm>

 #include <vector>

 #include <string>

 #include <cstring>

 using namespace std;

 int main()

 {

     int N, tmpNum, tmpFloor = , sum = ;

     cin >> N;

     for(int i = ; i < N; ++ i)

     {

         cin >> tmpNum;

         if(tmpFloor < tmpNum)

             sum +=  + *(tmpNum - tmpFloor);

         else

             sum +=  + *(tmpFloor - tmpNum);

         tmpFloor = tmpNum;

     }

     cout << sum;

     return ;

 }

巴特西

PAT A1005-1008

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