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 ;
 }