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1039. Course List for Student (25)-PAT甲级真题

Zhejiang University has 40000 students and provides 2500 courses. Now given the student name lists of all the courses, you are supposed to output the registered course list for each student who comes for a query.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (<=40000), the number of students who look for their course lists, and K (<=2500), the total number of courses. Then the student name lists are given for the courses (numbered from 1 to K) in the following format: for each course i, first the course index i and the number of registered students Ni (<= 200) are given in a line. Then in the next line, Ni student names are given. A student name consists of 3 capital English letters plus a one-digit number. Finally the last line contains the N names of students who come for a query. All the names and numbers in a line are separated by a space.

Output Specification:

For each test case, print your results in N lines. Each line corresponds to one student, in the following format: first print the student’s name, then the total number of registered courses of that student, and finally the indices of the courses in increasing order. The query results must be printed in the same order as input. All the data in a line must be separated by a space, with no extra space at the end of the line.

Sample Input:
11 5
4 7
BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
1 4
ANN0 BOB5 JAY9 LOR6
2 7
ANN0 BOB5 FRA8 JAY9 JOE4 KAT3 LOR6
3 1
BOB5
5 9
AMY7 ANN0 BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
ZOE1 ANN0 BOB5 JOE4 JAY9 FRA8 DON2 AMY7 KAT3 LOR6 NON9
Sample Output:
ZOE1 2 4 5
ANN0 3 1 2 5
BOB5 5 1 2 3 4 5
JOE4 1 2
JAY9 4 1 2 4 5
FRA8 3 2 4 5
DON2 2 4 5
AMY7 1 5
KAT3 3 2 4 5
LOR6 4 1 2 4 5
NON9 0
题目大意：有N个学生，K门课，给出选择每门课的学生姓名，最后对于给出的N个学生的姓名的选课情况进行询问，要求按顺序给出每个学生的选课情况～
分析：考虑到string、cin、cout会超时，可以使用hash(26*26*26*10+10)将学生姓名变为int型，然后存储在vector里面～～

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int getid(char *name) {
    int id = 0;
    for(int i = 0; i < 3; i++)
        id = 26 * id + (name[i] - 'A');
    id = id * 10 + (name[3] - '0');
    return id;
}
const int maxn = 26 * 26 * 26 * 10 + 10;
vector<int> v[maxn];

int main() {
    int n, k, no, num, id = 0;
    char name[5];
    scanf("%d %d", &n, &k);
    for(int i = 0; i < k; i++) {
        scanf("%d %d", &no, &num);
        for(int j = 0; j < num; j++) {
            scanf("%s", name);
            id = getid(name);
            v[id].push_back(no);
        }
    }
    for(int i = 0; i < n; i++) {
        scanf("%s", name);
        id = getid(name);
        sort(v[id].begin(), v[id].end());
        printf("%s %lu", name, v[id].size());
        for(int j = 0; j < v[id].size(); j++)
            printf(" %d", v[id][j]);
        printf("\n");
    }
    return 0;
}

#include <cstdio>

#include <vector>

#include <algorithm>

using namespace std;

int getid(char *name) {

int id = 0;

for(int i = 0; i < 3; i++)

id = 26 * id + (name[i] - 'A');

id = id * 10 + (name[3] - '0');

return id;

}

const int maxn = 26 * 26 * 26 * 10 + 10;

vector<int> v[maxn];

int main() {

int n, k, no, num, id = 0;

char name[5];

scanf("%d %d", &n, &k);

for(int i = 0; i < k; i++) {

scanf("%d %d", &no, &num);

for(int j = 0; j < num; j++) {

scanf("%s", name);

id = getid(name);

v[id].push_back(no);

}

for(int i = 0; i < n; i++) {

scanf("%s", name);

id = getid(name);

sort(v[id].begin(), v[id].end());

printf("%s %lu", name, v[id].size());

for(int j = 0; j < v[id].size(); j++)

printf(" %d", v[id][j]);

printf("\n");

}

return 0;

}

1048. Find Coins (25)-PAT甲级真题(Hash散列)

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she could only use exactly two coins to pay the exact amount. Since she has as many as 105 coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find two coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (<=105, the total number of coins) and M(<=103, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers no more than 500. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the two face values V1 and V2 (separated by a space) such that V1 + V2 = M and V1 <= V2. If such a solution is not unique, output the one with the smallest V1. If there is no solution, output “No Solution” instead.

Sample Input 1:
8 15
1 2 8 7 2 4 11 15
Sample Output 1:
4 11
Sample Input 2:
7 14
1 8 7 2 4 11 15
Sample Output 2:
No Solution
题目大意：给出n个正整数和一个正整数m，问n个数字中是否存在一对数字a和b(a <= b),使a+b=m成立。如果有多个，输出a最小的那一对。
分析：建立数组a保存每个数字出现的次数，然后判断输出

#include <iostream>
using namespace std;
int a[1001];
int main() {
    int n, m, temp;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < n; i++) {
        scanf("%d", &temp);
        a[temp]++;
    }
    for(int i = 0; i < 1001; i++) {
        if(a[i]) {
            a[i]--;
            if(m > i && a[m-i]) {
                printf("%d %d", i, m - i);
                return 0;
            }
            a[i]++;
        }
    }
    printf("No Solution");
    return 0;
}

#include <iostream>

using namespace std;

int a[1001];

int main() {

int n, m, temp;

scanf("%d %d", &n, &m);

for(int i = 0; i < n; i++) {

scanf("%d", &temp);

a[temp]++;

}

for(int i = 0; i < 1001; i++) {

if(a[i]) {

a[i]--;

if(m > i && a[m-i]) {

printf("%d %d", i, m - i);

return 0;

}

a[i]++;

}

printf("No Solution");

return 0;

}

1041. Be Unique (20)-PAT甲级真题（Hash散列）

Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1, 104]. The first one who bets on a unique number wins. For example, if there are 7 people betting on 5 31 5 88 67 88 17, then the second one who bets on 31 wins.

Input Specification:

Each input file contains one test case. Each case contains a line which begins with a positive integer N (<=105) and then followed by N bets. The numbers are separated by a space.

Output Specification:

For each test case, print the winning number in a line. If there is no winner, print “None” instead.

Sample Input 1:
7 5 31 5 88 67 88 17
Sample Output 1:
31
Sample Input 2:
5 888 666 666 888 888
Sample Output 2:
None
题目大意：给n个数字，按照读入顺序，哪个数字是第一个在所有数字中只出现一次的数字。如果所有数字出现都超过了一次，则输出None
分析：建立一个数组，存储每个数字出现的次数，然后遍历一遍输入的顺序看是否有出现次数为1的数字

#include <cstdio>
using namespace std;
int a[100001], m[100000];
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
        m[a[i]]++;
    }
    for(int i = 0; i < n; i++) {
        if(m[a[i]] == 1) {
            printf("%d", a[i]);
            return 0;
        }
    }
    printf("None");
    return 0;
}

#include <cstdio>

using namespace std;

int a[100001], m[100000];

int main() {

int n;

scanf("%d", &n);

for(int i = 0; i < n; i++) {

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

m[a[i]]++;

}

for(int i = 0; i < n; i++) {

if(m[a[i]] == 1) {

printf("%d", a[i]);

return 0;

}

printf("None");

return 0;

}

cin.getline()、getline()、gets()、cin.get()，getchar()的区别

此文献给风临神上。

#include <iostream>
string str1;
char str2[100];
cin >> str1;
cin >> str2;//cin以空格和回车为结束符

char ch;
cin.get(ch);//用来接收一个字符

char str[20];
cin.get(str, 20);//接收一行字符串，可以接收空格

char str[20];
cin.getline(str, 20);//接收一行字符串，可以接收空格
cin.getline(str, 20, '#');//接收一行字符串，可以接收空格，判断直到遇到'#'为止

#include <iostream>

string str1;

char str2[100];

cin >> str1;

cin >> str2;//cin以空格和回车为结束符

char ch;

cin.get(ch);//用来接收一个字符

char str[20];

cin.get(str, 20);//接收一行字符串，可以接收空格

char str[20];

cin.getline(str, 20);//接收一行字符串，可以接收空格

cin.getline(str, 20, '#');//接收一行字符串，可以接收空格，判断直到遇到'#'为止

#include <string>
string s;
getline(cin, s);//接收一行字符串
//getline()和cin.getline()很类似，cin.getline()属于istream流，需要包含头文件<iostream>
//而getline()属于string流，需要包含头文件<string>,是不一样的两个函数

char str[20];
gets(str);//接收一行字符串到字符数组里面
//gets使用的时候编译器可能会产生警告：warning: this program uses gets(), which is unsafe.  所以建议使用cin.getline()

char ch;
ch = getchar();//不能写成getchar(ch);

#include <string>

string s;

getline(cin, s);//接收一行字符串

//getline()和cin.getline()很类似，cin.getline()属于istream流，需要包含头文件<iostream>

//而getline()属于string流，需要包含头文件<string>,是不一样的两个函数

char str[20];

gets(str);//接收一行字符串到字符数组里面

//gets使用的时候编译器可能会产生警告：warning: this program uses gets(), which is unsafe. 所以建议使用cin.getline()

char ch;

ch = getchar();//不能写成getchar(ch);

1050. String Subtraction (20)-PAT甲级真题（哈希）

Given two strings S1 and S2, S = S1 – S2 is defined to be the remaining string after taking all the characters in S2 from S1. Your task is simply to calculate S1 – S2 for any given strings. However, it might not be that simple to do it fast.

Input Specification:

Each input file contains one test case. Each case consists of two lines which gives S1 and S2, respectively. The string lengths of both strings are no more than 104. It is guaranteed that all the characters are visible ASCII codes and white space, and a new line character signals the end of a string.

Output Specification:

For each test case, print S1 – S2 in one line.

Sample Input:
They are students.
aeiou
Sample Output:
Thy r stdnts.
题目大意：给出两个字符串，在第一个字符串中删除第二个字符串中出现过的所有字符并输出～
分析：用flag[256]数组变量标记str2出现过的字符为true，输出str1的时候根据flag[str1[i]]是否为true,如果是true就不输出
注意：使用int lens1 = strlen(s1);int lens2 = strlen(s2);的形式，否则直接放在for循环里面会超时～
如果使用gets前面请使用char str[100000]而非char *str～～

#include <iostream>
#include <string>
#include <string.h>
using namespace std;
char s1[100000], s2[100000];
int main() {
    cin.getline(s1, 100000);
    cin.getline(s2, 100000);
    int lens1 = strlen(s1), lens2 = strlen(s2);
    bool flag[256] = {false};
    for(int i = 0; i < lens2; i++)
        flag[s2[i]] = true;
    for(int i = 0; i < lens1; i++) {
        if(!flag[s1[i]])
            printf("%c", s1[i]);
    }
    return 0;
}

#include <iostream>

#include <string>

#include <string.h>

using namespace std;

char s1[100000], s2[100000];

int main() {

cin.getline(s1, 100000);

cin.getline(s2, 100000);

int lens1 = strlen(s1), lens2 = strlen(s2);

bool flag[256] = {false};

for(int i = 0; i < lens2; i++)

flag[s2[i]] = true;

for(int i = 0; i < lens1; i++) {

if(!flag[s1[i]])

printf("%c", s1[i]);

}

return 0;

}

1083. List Grades (25)-PAT甲级真题

Given a list of N student records with name, ID and grade. You are supposed to sort the records with respect to the grade in non-increasing order, and output those student records of which the grades are in a given interval.

Input Specification:

Each input file contains one test case. Each case is given in the following format:

N
name[1] ID[1] grade[1]
name[2] ID[2] grade[2]
… …
name[N] ID[N] grade[N]
grade1 grade2
where name[i] and ID[i] are strings of no more than 10 characters with no space, grade[i] is an integer in [0, 100], grade1 and grade2 are the boundaries of the grade’s interval. It is guaranteed that all the grades are distinct.

Output Specification:

For each test case you should output the student records of which the grades are in the given interval [grade1, grade2] and are in non-increasing order. Each student record occupies a line with the student’s name and ID, separated by one space. If there is no student’s grade in that interval, output “NONE” instead.

Sample Input 1:
4
Tom CS000001 59
Joe Math990112 89
Mike CS991301 100
Mary EE990830 95
60 100
Sample Output 1:
Mike CS991301
Mary EE990830
Joe Math990112
Sample Input 2:
2
Jean AA980920 60
Ann CS01 80
90 95
Sample Output 2:
NONE

题目大意：给出n个考生的信息，按照分数从高到低排序，并且输出给定区间的考生信息。如果不存在满足条件的考生就输出NONE。
分析：建立结构体数组，将不满足条件的学生grade改为-1，并统计满足条件的学生的个数cnt，然后进行排序，输出前cnt个考生的信息～～
注意：因为每个学生的成绩都不同，所以按照下降排列即可，return a.grade > b.grade;

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct stu {
    char name[12];
    char id[12];
    int grade;
};
int cmp1(stu a, stu b) {
    return a.grade > b.grade;
}
int main() {
    int n, low, high, cnt = 0;
    scanf("%d", &n);
    vector<stu> v(n);
    for(int i = 0; i < n; i++) {
        scanf("%s %s %d", v[i].name, v[i].id, &v[i].grade);
    }
    scanf("%d %d", &low, &high);
    for(int i = 0; i < n; i++) {
        if(v[i].grade < low || v[i].grade > high) {
            v[i].grade = -1;
        } else {
            cnt++;
        }
    }
    sort(v.begin(), v.end(), cmp1);
    for(int i = 0; i < cnt; i++) {
        printf("%s %s\n", v[i].name, v[i].id);
    }
    if(cnt == 0)
        printf("NONE");
    return 0;
}

#include <iostream>

#include <vector>

#include <algorithm>

using namespace std;

struct stu {

char name[12];

char id[12];

int grade;

};

int cmp1(stu a, stu b) {

return a.grade > b.grade;

}

int main() {

int n, low, high, cnt = 0;

scanf("%d", &n);

vector<stu> v(n);

for(int i = 0; i < n; i++) {

scanf("%s %s %d", v[i].name, v[i].id, &v[i].grade);

}

scanf("%d %d", &low, &high);

for(int i = 0; i < n; i++) {

if(v[i].grade < low || v[i].grade > high) {

v[i].grade = -1;

} else {

cnt++;

}

sort(v.begin(), v.end(), cmp1);

for(int i = 0; i < cnt; i++) {

printf("%s %s\n", v[i].name, v[i].id);

}

if(cnt == 0)

printf("NONE");

return 0;

}