PAT A1015-1016

A 1015 Reversible Primes

　　看清题意即可。给的数是十进制的，需要先判断是不是素数，然后按照给定进制转化成字符串后进行翻转，最后再转化为十进制并判断是否为素数。

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <string>

 #include <cstdlib>

 #include <cmath>

 #include <cstring>

 using namespace std;

 int N, D;

 char numStr[];

 bool isPrime(int tmpNum)

 {

     if(tmpNum < )

         return false;

     int maxNum = sqrt(1.0*tmpNum);

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

         if(tmpNum % i == )

             return false;

     return true;

 }

 int strToNum()

 {

     int len = strlen(numStr), tmpNum = ;

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

     {

         tmpNum = tmpNum*D + numStr[i]-'';

     }

     return tmpNum;

 }

 int main()

 {

     int tmpNum;

     while(scanf("%d%d", &N, &D) != EOF && N >= )

     {

         if(!isPrime(N))

         {

             cout << "No" << endl;

             continue;

         }

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

         {

             numStr[i] = N % D + '';

             N /= D;

             numStr[i+] = '\0';

         }

         tmpNum = strToNum();

         if(!isPrime(tmpNum))

             cout << "No" << endl;

         else

             cout << "Yes" << endl;

     }

     return ;

 }

A 1016 1016 Phone Bills

　因为存在无效的记录，所以需要将他们先进行配对。处理方法是：按照名字和时间进行排序，然后筛选出所需要的记录再进行下一步处理。

　超时问题：每一个有效时间段的话费计算需要尽量减少计算量。

以下是柳婼精简的代码：

 #include <iostream>

 #include <map>

 #include <vector>

 #include <algorithm>

 using namespace std;

 struct node {

     string name;

     int status, month, time, day, hour, minute;

 };

 bool cmp(node a, node b) {

     return a.name != b.name ? a.name < b.name : a.time < b.time;

 }

 double billFromZero(node call, int *rate) {

     double total = rate[call.hour] * call.minute + rate[] *  * call.day;

     for (int i = ; i < call.hour; i++)

         total += rate[i] * ;

     return total / 100.0;

 }

 int main() {

     int rate[] = {}, n;

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

         scanf("%d", &rate[i]);

         rate[] += rate[i];

     }

     scanf("%d", &n);

     vector<node> data(n);

     for (int i = ; i < n; i++) {

         cin >> data[i].name;

         scanf("%d:%d:%d:%d", &data[i].month, &data[i].day, &data[i].hour, &data[i].minute);

         string temp;

         cin >> temp;

         data[i].status = (temp == "on-line") ?  : ;

         data[i].time = data[i].day *  *  + data[i].hour *  + data[i].minute;

     }

     sort(data.begin(), data.end(), cmp);

     map<string, vector<node> > custom;

     for (int i = ; i < n; i++) {

         if (data[i].name == data[i - ].name && data[i - ].status ==  && data[i].status == ) {

             custom[data[i - ].name].push_back(data[i - ]);

             custom[data[i].name].push_back(data[i]);

         }

     }

     for (auto it : custom) {

         vector<node> temp = it.second;

         cout << it.first;

         printf(" %02d\n", temp[].month);

         double total = 0.0;

         for (int i = ; i < temp.size(); i += ) {

             double t = billFromZero(temp[i], rate) - billFromZero(temp[i - ], rate);

             printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n", temp[i - ].day, temp[i - ].hour, temp[i - ].minute, temp[i].day, temp[i].hour, temp[i].minute, temp[i].time - temp[i - ].time, t);

             total += t;

         }

         printf("Total amount: $%.2f\n", total);

     }

     return ;

 }

巴特西

PAT A1015-1016

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