PAT A1009-1012

A 1009 Product of Polynomials (25 point(s))

　　读懂题意就行。

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

 using namespace std;
 typedef struct NODE
 {
     int exp;
     double coe;
     NODE(){exp = ; coe = ;}
     NODE(int e, double v):exp(e),coe(v){}
 }node;
 vector<node> poly1, poly2;
 double rstPoly[]={};
 int main()
 {
     int N, tmpExp;
     double tmpCoe;
     cin >> N;
     for(int i = ; i < N; ++ i)
     {
         cin >> tmpExp >> tmpCoe;
         poly1.push_back(NODE(tmpExp, tmpCoe));
     }
     cin >> N;
     for(int i = ; i < N; ++ i)
     {
         cin >> tmpExp >> tmpCoe;
         poly2.push_back(NODE(tmpExp, tmpCoe));
     }
     for(int i = ; i < poly1.size(); ++i)
         for(int j = ; j < poly2.size(); ++ j)
             rstPoly[poly1[i].exp+poly2[j].exp] += poly1[i].coe*poly2[j].coe;
     int cnt = ;
     for(int i = ; i < ; ++ i)
         if(rstPoly[i] != )
             cnt++;
     cout << cnt;
     for(int i = ; i >= ; -- i)
         if(rstPoly[i] != )
         {
             printf(" %d %.1f", i, rstPoly[i]);
         }
     return ;
 }

A 1010 Radix (25 point(s))

　　这道题目相当有趣，坑的莫名其妙，哈哈哈。

　　注意：1.数的范围，10位数字，int 不够用

　　　　　2.超时，因而不能从小到大递增取进制数

　　　　   3.二分法，最大进制和最小进制的确定

　　　　   4.超限，在找进制数的过程中，过大的进制数会发生超限，注意这种情况

　　　　　5.待确定进制数只有一位，此时需要注意取最小进制

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <cstdlib>
 #include <vector>
 #include <string>
 #include <algorithm>

 using namespace std;
 long long toNum(char c)
 {
     return isdigit(c) ? c-'' : c-'a'+;
 }
 long long strRadixToNum(string tmpStr, long long radix)
 {
     long long tmpNum = , ant = ;
     for(int i = tmpStr.size()-; i >= ; -- i)
     {
         tmpNum += toNum(tmpStr[i])*ant;
         if(tmpNum <  || ant < )
             return -;
         ant *= radix;
     }
     return tmpNum;
 }
 long long minRadix(string tmpStr)
 {
     char tmpChar = ;
     for(int i = ; i < tmpStr.size(); ++ i)
     {
         if(tmpStr[i] > tmpChar)
             tmpChar = tmpStr[i];
     }
     return toNum(tmpChar) + ;
 }
 int main()
 {
     long long tag, radix, tmpNum, tmpTarget, lowRadix, highRadix;
     string N1, N2;
     cin >> N1 >> N2 >> tag >> radix;
     //算出进制已经确定的数作为目标值
     //并且 找出未确定进制的数的最小进制
     if(tag == )
         swap(N1, N2);
     highRadix = tmpTarget = strRadixToNum(N1, radix);
     lowRadix = max((long long) , minRadix(N2));
     while(lowRadix <= highRadix)
     {
         long long midRadix = (lowRadix + highRadix)/;
         long long tmpNum1 = strRadixToNum(N2, midRadix);
         if(tmpNum1 >= tmpTarget ||tmpNum1 == -)
              highRadix = midRadix - ;
         else
             lowRadix = midRadix + ;
     }
     if(strRadixToNum(N2, lowRadix) == tmpTarget)
         cout << lowRadix;
     else
         cout << "Impossible";
     return ;
 }

A 1011 World Cup Betting (20 point(s))

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <cstdlib>
 #include <vector>
 #include <algorithm>

 using namespace std;
 int main()
 {
     double sum = 0.65, tmpW, tmpT, tmpL;
     for(int i = ; i < ; ++ i)
     {
         cin >> tmpW >> tmpT >> tmpL;
         if(tmpW > max(tmpT, tmpL))
         {
             cout << "W ";
             sum *= tmpW;
         }
         else if(tmpT > max(tmpW, tmpL))
         {
             cout << "T ";
             sum *= tmpT;
         }
         else
         {
             cout << "L ";
             sum *= tmpL;;
         }
     }
     printf("%.2f", (sum-)*);
     return ;
 }

A 1012 The Best Rank (25 point(s))

　　注意：1.排名的问题，相同成绩同一排名

　　　　　2.最佳排名相同，按所给优先级

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <cstdlib>
 #include <vector>
 #include <algorithm>
 #include <set>
 #include <unordered_map>
 #include <functional>

 using namespace std;
 typedef struct NODE
 {
     int id, math, cCode, aver, english;
 }node;
 vector<int> cCodeVec, mathVec, engVec, averVec;
 unordered_map<int, node> nodeMap;
 unordered_map<int, int> nodeTableMap;
 void getBestRank(int u)
 {
     int type = , rank, cnt;
     char typeStr[]="ACME";
     node tmpNode = nodeMap[u];
     cnt = ;
     for(auto it = averVec.begin(); it != averVec.end(); ++it, ++cnt)
         if(*it == tmpNode.aver)
         {
             type = ; rank = cnt;break;
         }
     cnt = ;
     for(auto it = cCodeVec.begin(); it != cCodeVec.end(); ++it, ++cnt)
         if(*it == tmpNode.cCode && cnt < rank)
         {
             type = ; rank = cnt;break;
         }
     cnt = ;
     for(auto it = mathVec.begin(); it != mathVec.end(); ++it, ++cnt)
         if(*it == tmpNode.math && cnt < rank)
         {
             type = ; rank = cnt;break;
         }
     cnt = ;
     for(auto it = engVec.begin(); it != engVec.end(); ++it, ++cnt)
         if(*it == tmpNode.english && cnt < rank)
         {
             type = ; rank = cnt;break;
         }
     printf("%d %c\n", rank, typeStr[type]);
 }
 int main()
 {
     int N, M;
     node tmpNode;
     cin >> N >> M;
     for(int i = ; i < N; ++i)
     {
         cin >> tmpNode.id >> tmpNode.cCode >> tmpNode.math >> tmpNode.english;
         tmpNode.aver = (tmpNode.cCode + tmpNode.math + tmpNode.english+1.5)/;
         cCodeVec.push_back(tmpNode.cCode); mathVec.push_back(tmpNode.math);
         engVec.push_back(tmpNode.english); averVec.push_back(tmpNode.aver);
         nodeMap[tmpNode.id] = tmpNode;nodeTableMap[tmpNode.id] = ;
     }
     sort(averVec.begin(), averVec.end(), greater<int>());
     sort(cCodeVec.begin(), cCodeVec.end(), greater<int>());
     sort(mathVec.begin(), mathVec.end(), greater<int>());
     sort(engVec.begin(), engVec.end(), greater<int>());
     for(int i = ; i < M; ++ i)
     {
         cin >> tmpNode.id;
         if(nodeTableMap[tmpNode.id] > )
             getBestRank(tmpNode.id);
         else
             cout << "N/A" << endl;
     }
     return ;
 }