PAT/查找元素习题集

B1004. 成绩排名 (20)

Description：

读入n名学生的姓名、学号、成绩，分别输出成绩最高和成绩最低学生的姓名和学号。

Input：

每个测试输入包含1个测试用例，格式为：

第1行：正整数n
  第2行：第1个学生的姓名 学号 成绩
  第3行：第2个学生的姓名 学号 成绩
  ... ... ...
  第n+1行：第n个学生的姓名 学号 成绩

其中姓名和学号均为不超过10个字符的字符串，成绩为0到100之间的一个整数，这里保证在一组测试用例中没有两个学生的成绩是相同的。

Output：

对每个测试用例输出2行，第1行是成绩最高学生的姓名和学号，第2行是成绩最低学生的姓名和学号，字符串间有1空格。

Sample Input：

3
Joe Math990112 89
Mike CS991301 100
Mary EE990830 95

Sample Output：

Mike CS991301
Joe Math990112

 #include <cstdio>

 struct Student{
     char name[];
     char id[];
     int score;
 }temp, ans_max, ans_min;

 int main()
 {
     int n;
     scanf("%d", &n);
     ans_max.score = -;
     ans_min.score = ;
     for(int i=; i<n; ++i) {
         scanf("%s%s%d", temp.name, temp.id, &temp.score);
         if(temp.score > ans_max.score)  ans_max = temp;
         if(temp.score < ans_min.score)  ans_min = temp;
     }

     printf("%s %s\n%s %s\n", ans_max.name, ans_max.id, ans_min.name, ans_min.id);

     return ;
 }

B1028. 人口普查 (20)

Description：

某城镇进行人口普查，得到了全体居民的生日。现请你写个程序，找出镇上最年长和最年轻的人。

这里确保每个输入的日期都是合法的，但不一定是合理的——假设已知镇上没有超过200岁的老人，而今天是2014年9月6日，所以超过200岁的生日和未出生的生日都是不合理的，应该被过滤掉。

Input：

输入在第一行给出正整数N，取值在(0, 10⁵]；随后N行，每行给出1个人的姓名（由不超过5个英文字母组成的字符串）、以及按“yyyy/mm/dd”（即年/月/日）格式给出的生日。题目保证最年长和最年轻的人没有并列。

Output：

在一行中顺序输出有效生日的个数、最年长人和最年轻人的姓名，其间以空格分隔。

Sample Input：

5
John 2001/05/12
Tom 1814/09/06
Ann 2121/01/30
James 1814/09/05
Steve 1967/11/20

Sample Output：

3 Tom John

 #include <cstdio>
 #include <cstring>

 #define MaxSize 11
 struct birthday {
     char name[MaxSize];
     char date[MaxSize];
 }temp, maxn, minn;
 char floor[MaxSize] = "1814/09/06", upper[MaxSize] = "2014/09/06";

 int main()
 {
     //freopen("E:\\Temp\\input.txt", "r", stdin);

     for(int i=; i<MaxSize; ++i) {
         maxn.date[i] = floor[i];
         minn.date[i] = upper[i];
     }
     int N;
     scanf("%d", &N);
     int counter = N;
     for(int i=; i<N; ++i) {
         scanf("%s %s", temp.name, temp.date);
         if(strcmp(temp.date, floor)< || strcmp(temp.date, upper)>)
             --counter;
         else {
             if(strcmp(temp.date, maxn.date) >= )   maxn = temp;
             if(strcmp(temp.date, minn.date) <= )   minn = temp;
         }
     }

     if(counter != )
         printf("%d %s %s\n", counter, minn.name, maxn.name);
     else
         printf("0\n");

     return ;
 }

 #include <cstdio>

 struct person {
     char name[];
     int yy, mm, dd;
 }oldest, youngest, left, right, temp;

 bool LessEqu(person a, person b)
 {
     if(a.yy != b. yy)   return a.yy <= b.yy;
     else if(a.mm != b.mm)   return a.mm <= b.mm;
     else return a.dd <= b.dd;
 }
 bool MoreEqu(person a, person b)
 {
     if(a.yy != b. yy)   return a.yy >= b.yy;
     else if(a.mm != b.mm)   return a.mm >= b.mm;
     else return a.dd >= b.dd;
 }
 void init()
 {
     youngest.yy = left.yy = ;
     oldest.yy = right.yy = ;
     youngest.mm = oldest.mm = left.mm = right.mm = ;
     youngest.dd = oldest.dd = left.dd = right.dd = ;
 }

 int main()
 {
     init();
     int n, num = ;
     scanf("%d", &n);
     for(int i=; i<n; ++i) {
         scanf("%s %d/%d/%d", temp.name, &temp.yy, &temp.mm, &temp.dd);
         if(MoreEqu(temp, left) && LessEqu(temp, right)) {
             num++;
             if(LessEqu(temp, oldest))  oldest = temp;
             if(MoreEqu(temp, youngest)) youngest = temp;
         }
     }

     if(num == )    printf("0\n");
     else printf("%d %s %s\n", num, oldest.name, youngest.name);

     return ;
 }

B1032. 挖掘机技术哪家强 (20)

Description：

为了用事实说明挖掘机技术到底哪家强，PAT组织了一场挖掘机技能大赛。现请你根据比赛结果统计出技术最强的那个学校。

Input：

输入在第1行给出不超过10⁵的正整数N，即参赛人数。随后N行，每行给出一位参赛者的信息和成绩，包括其所代表的学校的编号（从1开始连续编号）、及其比赛成绩（百分制），中间以空格分隔。

Output：

在一行中给出总得分最高的学校的编号、及其总分，中间以空格分隔。题目保证答案唯一，没有并列。

Sample Input：

6
3 65
2 80
1 100
2 70
3 40
3 0

Sample Output：

2 150

 #include <cstdio>

 const int maxn = ;
 int school[maxn];

 int main()
 {
     int n, schID, score;
     scanf("%d", &n);
     for(int i=; i<n; ++i) {
         scanf("%d%d", &schID, &score);
         school[schID] += score;
     }

     int k = , MAX = -;
     for(int i=; i<=n; ++i) {
         if(school[i] > MAX) {
             MAX = school[i];
             k = i;
         }
     }

     printf("%d %d\n", k, MAX);

     return ;
 }

A1011. World Cup Betting (20)

Description：

With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excited as the best players from the best teams doing battles for the World Cup trophy in South Africa. Similarly, football betting fans were putting their money where their mouths were, by laying all manner of World Cup bets.

Chinese Football Lottery provided a "Triple Winning" game. The rule of winning was simple: first select any three of the games. Then for each selected game, bet on one of the three possible results -- namely W for win, T for tie, and L for lose. There was an odd assigned to each result. The winner's odd would be the product of the three odds times 65%.

For example, 3 games' odds are given as the following:

 W    T    L
1.1  2.5  1.7
1.2  3.0  1.6
4.1  1.2  1.1

To obtain the maximum profit, one must buy W for the 3rd game, T for the 2nd game, and T for the 1st game. If each bet takes 2 yuans, then the maximum profit would be (4.1*3.0*2.5*65%-1)*2 = 37.98 yuans (accurate up to 2 decimal places).

Input：

Each input file contains one test case. Each case contains the betting information of 3 games. Each game occupies a line with three distinct odds corresponding to W, T and L.

Output：

For each test case, print in one line the best bet of each game, and the maximum profit accurate up to 2 decimal places. The characters and the number must be separated by one space.

Sample Input：

1.1 2.5 1.7
1.2 3.0 1.6
4.1 1.2 1.1

Sample Output：

T T W 37.98

 #include <cstdio>

 char S[] = {'W', 'T', 'L'};

 int main()
 {
     double ans = , temp, a;
     int idx;
     for(int i=; i<; ++i) {
         temp = ;
         for(int j=; j<; ++j) {
             scanf("%lf", &a);
             if(a > temp) {
                 temp = a;
                 idx = j;
             }
         }
         ans *= temp;
         printf("%c ", S[idx]);
     }

     printf("%.2f\n", (ans*0.65-)*);

     return ;
 }

A1006. Sign In and Sign Out (25)

Description：

At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to find the ones who have unlocked and locked the door on that day.

Input：

Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M, which is the total number of records, followed by M lines, each in the format:

ID_number Sign_in_time Sign_out_time

where times are given in the format HH:MM:SS, and ID number is a string with no more than 15 characters.

Output：

For each test case, output in one line the ID numbers of the persons who have unlocked and locked the door on that day. The two ID numbers must be separated by one space.

Note: It is guaranteed that the records are consistent. That is, the sign in time must be earlier than the sign out time for each person, and there are no two persons sign in or out at the same moment.

Sample Input：

3
CS301111 15:30:28 17:00:10
SC3021234 08:00:00 11:25:25
CS301133 21:45:00 21:58:40

Sample Output：

SC3021234 CS301133

 #include <cstdio>
 #include <cstring>

 #define MaxSize 20
 struct Person {
     char id[MaxSize];
     char in[MaxSize];
     char out[MaxSize];
 }temp, In, Out;

 int main()
 {
     //freopen("E:\\Temp\\input.txt", "r", stdin);

     char a[] = "00:00:00", b[] = "23:59:59";
     for(int i=; i<; ++i) {
         In.in[i] = Out.in[i]  = b[i];
         In.out[i] = Out.out[i]  = a[i];
     }
     int N;
     scanf("%d", &N);
     for(int i=; i<N; ++i) {
         scanf("%s %s %s", temp.id, temp.in, temp.out);
         if(strcmp(temp.in, In.in) < ) In = temp;
         if(strcmp(temp.out, Out.out) > ) Out = temp;
     }

     printf("%s %s\n", In.id, Out.id);

     return ;
 }

 #include <cstdio>

 struct pNode {
     char id[];
     int hh, mm, ss;
 }temp, ans1, ans2;

 bool great(pNode node1, pNode node2) {
     if(node1.hh != node2.hh)    return node1.hh > node2.hh;
     if(node1.mm != node2.mm)    return node1.mm > node2.mm;
     return node1.ss > node2.ss;
 }

 int main()
 {
     int n;
     scanf("%d", &n);
     ans1.hh = , ans1.mm = , ans1.ss = ;
     ans2.hh = , ans2.mm = , ans2.ss = ;
     for(int i=; i<n; ++i) {
         scanf("%s %d:%d:%d", temp.id, &temp.hh, &temp.mm, &temp.ss);
         if(great(temp, ans1) == false)  ans1 = temp;
         scanf("%d:%d:%d", &temp.hh, &temp.mm, &temp.ss);
         if(great(temp, ans2) == true)   ans2 = temp;
     }

     printf("%s %s\n", ans1.id, ans2.id);

     return ;
 }

A1036. Boys vs Girls (25)

Description：

This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students.

Input：

Each input file contains one test case. Each case contains a positive integer N, followed by N lines of student information. Each line contains a student's name, gender, ID and grade, separated by a space, where name and ID are strings of no more than 10 characters with no space, gender is either F (female) or M (male), and grade is an integer between 0 and 100. It is guaranteed that all the grades are distinct.

Output：

For each test case, output in 3 lines. The first line gives the name and ID of the female student with the highest grade, and the second line gives that of the male student with the lowest grade. The third line gives the difference grade_F-grade_M. If one such kind of student is missing, output "Absent" in the corresponding line, and output "NA" in the third line instead.

Sample Input1：

3

Joe M Math990112 89

Mike M CS991301 100

Mary F EE990830 95

Sample Output1：

Mary EE990830

Joe Math990112

6

Sample Input 2：

1

Jean M AA980920 60

Sample Output 2：

Absent

Jean AA980920

NA

 #include <cstdio>

 struct person {
     char name[];
     char id[];
     int score;
 }M, F, temp;

 void init()
 {
     M.score = ;
     F.score = -;
 }

 int main()
 {
     init();
     int n;
     char gender;
     scanf("%d", &n);
     for(int i=; i<n; ++i) {
         scanf("%s %c %s %d", temp.name, &gender, temp.id, &temp.score);
         if(gender=='M' && temp.score<M.score) M = temp;
         else if(gender=='F' && temp.score>F.score)  F = temp;
     }

     if(F.score == -)   printf("Absent\n");
     else printf("%s %s\n", F.name, F.id);
     if(M.score == )  printf("Absent\n");
     else printf("%s %s\n", M.name, M.id);
     if(F.score==- || M.score==) printf("NA\n");
     else printf("%d\n", F.score-M.score);

     return ;
 }