PAT 1069 1070 1071 1072

pat 1069 The Black Hole of Numbers

水题，代码如下：

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<algorithm>
 using namespace std;

 bool isSame(char buf[])
 {
     int i = ;
     while(buf[i] != '\0')
     {
         if(buf[i] != buf[])return false;
         else i++;
     }
     return true;
 }

 int main()
 {
     int num;
     scanf("%d", &num);
     char buf[];
     sprintf(buf, "%04d", num);
     if(isSame(buf))
     {
         printf("%04d - %04d = 0000", num, num);
         return ;
     }
     int len = strlen(buf);
     do
     {
         sort(buf, buf+len, greater<char>());
         int decre = atoi(buf);
         sort(buf, buf+len);
         int incre = atoi(buf);
         num = decre - incre;
         sprintf(buf, "%04d", num);
         printf("%04d - %04d = %04d\n", decre, incre, num);
     }while(num != );
     return ;
 }

---

pat 1070 Mooncake

水题，按照单价贪心选择即可。注意的是题目中的月饼的存货量用double类型，用int第三个数据通不过。代码如下：

 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 using namespace std;

 struct UnitPrice
 {
     double val;
     int index;
 };

 bool comp(UnitPrice a, UnitPrice b)
 {
     return a.val > b.val;
 }

 int main()
 {
     //freopen("input.txt","r",stdin);
     int N,D;
     scanf("%d%d", &N, &D);
     double *amount = new double[N];
     double *price = new double[N];
     UnitPrice *unitPrice = new UnitPrice[N];
     for(int i = ; i < N; i++)
         scanf("%lf", &amount[i]);
     for(int i = ; i < N; i++)
     {
         scanf("%lf", &price[i]);
         unitPrice[i].val = price[i] / amount[i];
         unitPrice[i].index = i;
     }
     sort(&unitPrice[], &unitPrice[N], comp);
     double profit = 0.0;
     int j = ;
     while(D >  && j < N)
     {
         int k = unitPrice[j++].index;
         if(D >= amount[k])
         {
             profit += price[k];
             D -= amount[k];
         }
         else
         {
             profit += (D*1.0/amount[k])*price[k];
             D = ;
         }
     }
     printf("%.2f\n" ,profit);
     return ;
 }

---

pat 1071 Speech Patterns                                                  本文地址

水题，遍历字符串，统计每个单词出现次数即可。代码如下：

 #include<iostream>
 #include<string>
 #include<map>
 #include<algorithm>
 using namespace std;

 bool isValid(char c)
 {
     if((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') ||
        (c >= '' && c <= ''))
         return true;
     else return false;
 }

 int main()
 {
     string buf, maxWord;
     getline(cin, buf);
     transform(buf.begin(),buf.end(),buf.begin(), ::tolower);
     map<string,int> words;
     int i = , times = ;
     while( i < buf.length() )
     {
         while( i < buf.length() && isValid(buf[i]) == false) i++;
         if(i < buf.length())
         {
             int start = i++;
             while( i < buf.length() && isValid(buf[i]) == true) i++;
             if( i <= buf.length())// 这里注意要<=,不能是<
             {
                 string word = buf.substr(start, i-start);
                 if(words.find(word) == words.end())
                 {
                     words[word] = ;
                     if(times < ){times = ; maxWord = word;}
                 }
                 else
                 {
                     words[word] ++;
                     if(times < words[word] ||
                        (times == words[word] && word < maxWord))
                     {
                         times = words[word];
                         maxWord = word;
                     }
                 }
             }
         }
     }
     cout<<maxWord<<" "<<times;
     return ;
 }

---

pat 1072 Gas Station                                                         本文地址

题目的意思是选出一个加油站，求出加油站到几个house的距离的最小值，从这些最小值中选出一个最小的记为k(即距离station最近的的house)，选择的加油站要使k最大。理解题目意思后就很简单了，只要依次以每个候选加油站为起点计算单源最短路径，然后计算k，选择k最大的加油站即可。代码如下：

 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #include<climits>
 using namespace std;
 const int INF = INT_MAX;

 int index(char buf[], int houseNum)
 {
     if(buf[] == 'G')
         return atoi(buf+) + houseNum - ;
     else
         return atoi(buf) - ;
 }

 //以src为起点的单源最短路径
 void Dijkstra(int src, int dist[], int **graph, int n)
 {
     bool *used = new bool[n];
     for(int i = ; i < n; i++)
         {dist[i] = graph[src][i]; used[i] = false;}
     for(int i = ; i < n; i++)
     {
         int tmin = INF,k;
         for(int j = ; j < n; j++)
             if(!used[j] && tmin > dist[j])
             {
                 tmin = dist[j];
                 k = j;
             }
         used[k] = true;
         for(int j = ; j < n; j++)
             if(dist[k] != INF && graph[k][j] != INF &&
                dist[k] + graph[k][j] < dist[j])
             {
                 dist[j] = dist[k] + graph[k][j];
             }
     }
     delete used;
 }
 int main()
 {
     //freopen("input.txt", "r", stdin);
     int houseNum, stationNum,roadNum,range;
     scanf("%d%d%d%d", &houseNum, &stationNum, &roadNum, &range);
     const int nodeNum = houseNum + stationNum;
     int **graph = new int*[nodeNum];
     for(int i = ; i < nodeNum; i++)
     {
         graph[i] = new int[nodeNum];
         for(int j = ; j < nodeNum; j++)
             graph[i][j] = (i != j ? INF:);
     }
     for(int i = ; i < roadNum; i++)
     {
         char buf[];
         scanf("%s", buf); int a = index(buf, houseNum);
         scanf("%s", buf); int b = index(buf, houseNum);
         scanf("%d", &graph[a][b]);
         graph[b][a] = graph[a][b];
     }

     double minDistance = -1.0, averageDistance = INF;
     int Selectstation = -;
     int dist[nodeNum];
     for(int i = houseNum; i < nodeNum; i++)
     {
         Dijkstra(i, dist, graph, nodeNum);
         double mindis = INF, averdis = 0.0;
         bool flag = true;
         for(int j = ; j < houseNum; j++)
         {
             if(dist[j] > range){flag = false; break;}
             averdis += dist[j];
             if(dist[j] < mindis)mindis = dist[j];
         }
         if(flag == false)continue;
         averdis /= houseNum;
         if(mindis > minDistance)
         {
             minDistance = mindis;
             averageDistance = averdis;
             Selectstation = i;
         }
         else if(mindis == minDistance)
         {
             if(averdis < averageDistance)
             {
                 averageDistance = averdis;
                 Selectstation  = i;
             }
         }
     }
     if(Selectstation != -)
         printf("G%d\n%.1f %.1f\n", Selectstation+-houseNum, minDistance,
                 averageDistance);
     else printf("No Solution\n");
     return ;
 }

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