概述
PAT A1028 List Sorting 25分
- 原题
- 大体翻译
- 思路
- 第一次编写
- 结果
- 第二次编写
- 结果
- 第三次编写
- 结果
- 补充
原题
Excel can sort records according to any column. Now you are supposed to imitate this function.
Input Specification:
Each input file contains one test case. For each case, the first line contains two integers N (≤105 ) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then N lines follow, each contains a record of a student. A student’s record consists of his or her distinct ID (a 6-digit number), name (a string with no more than 8 characters without space), and grade (an integer between 0 and 100, inclusive).
Output Specification:
For each test case, output the sorting result in N lines. That is, if C = 1 then the records must be sorted in increasing order according to ID’s; if C = 2 then the records must be sorted in non-decreasing order according to names; and if C = 3 then the records must be sorted in non-decreasing order according to grades. If there are several students who have the same name or grade, they must be sorted according to their ID’s in increasing order.
Sample Input 1:
3 1
000007 James 85
000010 Amy 90
000001 Zoe 60
Sample Output 1:
000001 Zoe 60
000007 James 85
000010 Amy 90
Sample Input 2:
4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98
Sample Output 2:
000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60
Sample Input 3:
4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 90
Sample Output 3:
000001 Zoe 60
000007 James 85
000002 James 90
000010 Amy 90
大体翻译
类似于excel中的排序，排序的类型由第二个输入的数字C决定
当C为1时，则是根据第一列递增排序
当C为2时，则是根据第二列非递减排序
当C为3时，则是根据第三列非递减排序（注意审题）
每一份数据包括三个数据段，即学号，姓名和成绩
输入要排序的数据的数量N和上述所说的C
然后输入N个数据
最后输出排序完成后的结果即可
思路
一开始，用结构体存储数据，注意因为题目要求存储105个数据，所以需要定义成全局变量，否则空间会不够
然后根据C设置三个排序最后输出就可以了
第一次编程的时候使用了冒泡排序，以及cin，cout输入输出
第二次编程的时候使用了冒泡排序，以及scanf，printf输入输出
第三次编程的时候使用了sort函数，以及scanf，printf输入输出
第一次编写
#include <iostream>
using namespace std;
struct student
{
string id;
string name;
int cord;
}a[100000];
void sort(student a[], int n, int m)
{
student temp;
if (m == 1)
{
for (int i = 1; i <= n - 1; i++)
{
for (int j = 0; j < n - i; j++)
{
if (a[j].id > a[j + 1].id)
{
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
}
else if (m == 2)
{
for (int i = 1; i <= n - 1; i++)
{
for (int j = 0; j < n - i; j++)
{
if (a[j].name >= a[j + 1].name)
{
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
}
else if (m == 3)
{
for (int i = 1; i <= n - 1; i++)
{
for (int j = 0; j < n - i; j++)
{
if (a[j].cord >= a[j + 1].cord)
{
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
}
for (int i = 0; i < n; i++)
{
cout << a[i].id << " " << a[i].name << " " << a[i].cord << endl;
}
}
int main()
{
int n, c;
cin >> n >> c;
for (int i = 0; i < n; i++)
{
cin >> a[i].id >> a[i].name >> a[i].cord;
}
sort(a, n, c);
return 0;
}
结果
第二次编写
#include <iostream>
#include <stdio.h>
#include <cstring>
using namespace std;
struct student
{
char id[10];
char name[10];
int cord;
}a[100000];
void sort(student a[], int n, int m)
{
student temp;
if (m == 1)
{
for (int i = 1; i <= n - 1; i++)
{
for (int j = 0; j < n - i; j++)
{
if (strcmp(a[j].id, a[j + 1].id) > 0)
{
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
}
else if (m == 2)
{
for (int i = 1; i <= n - 1; i++)
{
for (int j = 0; j < n - i; j++)
{
if (strcmp(a[j].name, a[j + 1].name) >= 0)
{
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
}
else if (m == 3)
{
for (int i = 1; i <= n - 1; i++)
{
for (int j = 0; j < n - i; j++)
{
if (a[j].cord >= a[j + 1].cord)
{
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
}
for (int i = 0; i < n; i++)
{
//cout << a[i].id << " " << a[i].name << " " << a[i].cord << endl;
printf("%s %s %dn", a[i].id, a[i].name, a[i].cord);
}
}
int main()
{
int n, c;
cin >> n >> c;
for (int i = 0; i < n; i++)
{
//cin >> a[i].id >> a[i].name >> a[i].cord;
scanf("%s", a[i].id);
scanf("%s", a[i].name);
scanf("%d", &a[i].cord);
}
sort(a, n, c);
return 0;
}
结果
第三次编写
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
struct student
{
char id[10];
char name[10];
int cord;
}a[100000];
/*
void sort(student a[], int n, int m)
{
student temp;
if (m == 1)
{
for (int i = 1; i <= n - 1; i++)
{
for (int j = 0; j < n - i; j++)
{
if (strcmp(a[j].id, a[j + 1].id) > 0)
{
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
}
else if (m == 2)
{
for (int i = 1; i <= n - 1; i++)
{
for (int j = 0; j < n - i; j++)
{
if (strcmp(a[j].name, a[j + 1].name) >= 0)
{
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
}
else if (m == 3)
{
for (int i = 1; i <= n - 1; i++)
{
for (int j = 0; j < n - i; j++)
{
if (a[j].cord >= a[j + 1].cord)
{
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
}
for (int i = 0; i < n; i++)
{
//cout << a[i].id << " " << a[i].name << " " << a[i].cord << endl;
printf("%s %s %dn", a[i].id, a[i].name, a[i].cord);
}
}
*/
bool cmp1(student a, student b)
{
if (strcmp(a.id, b.id) < 0)
return 1;
else
return 0;
}
bool cmp2(student a, student b)
{
/*
if (strcmp(a.name, b.name) <= 0)
return 1;
else
return 0;
*/
if (strcmp(a.name, b.name) != 0)
return strcmp(a.name, b.name) < 0;
else
return strcmp(a.id,b.id) < 0;
}
bool cmp3(student a, student b)
{
//return a.cord <= b.cord;
if (a.cord != b.cord)
return a.cord < b.cord;
else
return strcmp(a.id, b.id) < 0;
}
int main()
{
int n, c;
cin >> n >> c;
for (int i = 0; i < n; i++)
{
//cin >> a[i].id >> a[i].name >> a[i].cord;
scanf("%s", a[i].id);
scanf("%s", a[i].name);
scanf("%d", &a[i].cord);
}
//sort(a, n, c);
switch (c)
{
case 1:sort(a, a + n, cmp1);
break;
case 2:sort(a, a + n, cmp2);
break;
case 3:sort(a, a + n, cmp3);
}
for (int i = 0; i < n; i++)
{
printf("%s %s %dn", a[i].id, a[i].name, a[i].cord);
}
return 0;
}
结果
补充
在最新的VS中要使用scanf_s而不能是scanf，但在提交中还是使用scanf
下面记录scanf_s的用法：
char a[10]; //字符串
scanf_s("%s",a,10); //需要输入变量的大小
int b; //整数
scanf_s("%d",&b,4);
下面记录最基础的sort函数的用法：
bool cmp(int a,int b)
{
return a<b; //注意大于小于
}
int main()
{
int a[10];
for(int i = 0; i < 10; i++)
scanf("%d",&a[i]);
sort(a, a + 10, cmp);
return 0;
}
题目中要求的是当姓名或分数一样，按照id进行排序，在第一次和第二次编写中，直接用了大于等于不对，但是不修改了
最后
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