题目链接:http://pat.zju.edu.cn/contests/pat-a-practise/1028

题目描述:

Excel can sort records according to any column. Now you are supposed to imitate this function.

Input

Each input file contains one test case. For each case, the first line contains two integers N (<=100000) 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

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

分析:

输入学生信息,按照指定项进行排序。当按照name或grade排序时如果存在一样的情况就按照学号递增的形式输出。

原本是想用c++做,num和name用string类型表示。但是用c++中的cin进行输入时会有一组数据在最终提交时超时。所以改用scanf输入。

由于scanf不能输入string类型,所以将string类型用char[]代替。

但是如果num和name都用char[]类型,在输出时又会出错,见“错误代码一”。

后面就将num为int型,为了满足num为6位的要求,要用printf("%06d",num) 这种形式输出。

错误代码一:

#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<vector>
#include<cstdio>
using namespace std; typedef struct student
{
    char num[6];
    char name[20];
    int grade;
}Student; int n;
bool comparison(Student a,Student b)
{
    if(n==1)
    {
        return strcmp(a.num,b.num)<0;
    }
    else if(n==2)
    {        
        if(strcmp(a.name,b.name) == 0)
            return strcmp(a.num,b.num);
        return strcmp(a.name,b.name)<0;
    }
    else if(n==3)
    {
        if(a.grade == b.grade)
            return strcmp(a.num,b.num) < 0;    
        return a.grade < b.grade;
    }        
} int main()
{
    int M;
    cin>>M>>n;
    vector<Student> s(M);
    int i;
    for(i=0; i<M; i++)
        //cin>>s[i].num>>s[i].name>>s[i].grade;
        scanf("%s%s%d",&s[i].num,&s[i].name,&s[i].grade);     sort(s.begin(),s.end(),comparison);
    for(i=0; i<M; i++)
        cout<<s[i].num<<" "<<s[i].name<<" "<<s[i].grade<<endl;     return 0;
}

最终的输出不正确。但是
原因是???


正确代码:

参考:http://blog.csdn.net/sunbaigui/article/details/8657115

#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<vector>
#include<cstdio>
using namespace std; typedef struct student
{
int num;
char name[20];
int grade;
}Student; int n;
bool comparison(Student a,Student b)
{
if(n==1)
{
return a.num<b.num;
}
else if(n==2)
{
if(strcmp(a.name,b.name) == 0)
return a.num<b.num;
return strcmp(a.name,b.name)<0;
}
else if(n==3)
{
if(a.grade == b.grade)
return a.num<b.num;
return a.grade < b.grade;
}
} int main()
{
int M;
cin>>M>>n;
vector<Student> s(M);
int i;
for(i=0; i<M; i++)
scanf("%d%s%d",&s[i].num,&s[i].name,&s[i].grade); sort(s.begin(),s.end(),comparison);
for(i=0; i<M; i++)
printf("%06d %s %d\n",s[i].num,s[i].name,s[i].grade); return 0;
}

【PAT】1028. List Sorting (25)的更多相关文章

  1. PAT 甲级 1028 List Sorting (25 分)(排序,简单题)

    1028 List Sorting (25 分)   Excel can sort records according to any column. Now you are supposed to i ...

  2. PAT 甲级 1028. List Sorting (25) 【结构体排序】

    题目链接 https://www.patest.cn/contests/pat-a-practise/1028 思路 就按照 它的三种方式 设计 comp 函数 然后快排就好了 但是 如果用 c++ ...

  3. 【PAT甲级】1028 List Sorting (25 分)

    题意: 输入一个正整数N(<=100000)和C(C属于{1,2,3}),接下来输入N行,每行包括学生的六位学号(习惯用string输入,因为可能有前导零),名字和成绩(正整数).输出排序后的信 ...

  4. 【PAT】1020 Tree Traversals (25)(25 分)

    1020 Tree Traversals (25)(25 分) Suppose that all the keys in a binary tree are distinct positive int ...

  5. 【PAT】1015 德才论 (25)(25 分)

    1015 德才论 (25)(25 分) 宋代史学家司马光在<资治通鉴>中有一段著名的“德才论”:“是故才德全尽谓之圣人,才德兼亡谓之愚人,德胜才谓之君子,才胜德谓之小人.凡取人之术,苟不得 ...

  6. 【PAT】1020. Tree Traversals (25)

    Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and i ...

  7. 【PAT】B1055 集体照(25 分)

    很简单的two points问题 ##注意:K是行数 #include<stdio.h> #include<string.h> #include<map> #inc ...

  8. 【PAT】1051 Pop Sequence (25)(25 分)

    Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and p ...

  9. 【PAT】1063. Set Similarity (25) 待改进

    Given two sets of integers, the similarity of the sets is defined to be Nc/Nt*100%, where Nc is the ...

随机推荐

  1. python解决组合问题

    1.问题描述 比如9个数中取4个数的组合以及列出各种组合,该如何做? 我们可以考虑以下一个简单组合:从1,2,3,4,5,6中,如何选取任意四个数的组合. 固定:1   2  3  ,组合有1234 ...

  2. [webpack] devtool里的7种SourceMap[转]

    modle: development cheap-source-map debug 不太方便,不是以原来的文件的形式cheap-module-source-map 可以 debugcheap-modu ...

  3. [CCC 2018] 平衡树

    题面在这里! 根据题目描述就可以直接模拟出一个暴力. 如果把前 n^(1/2) 的树的方案数先一遍 O(n^(3/4)) 暴力预处理出来(其实复杂度并到不了这个级别),然后把n带进来直接暴力算就行了. ...

  4. Codeforces Round #346 (Div. 2) G. Fence Divercity dp

    G. Fence Divercity 题目连接: http://www.codeforces.com/contest/659/problem/G Description Long ago, Vasil ...

  5. git fetch, git pull 以及 FETCH_HEAD

    git push. 这个很简单, 其实和后面的差不多, 这里就不讲了. 唯一需要注意的地方是: git push origin :branch2, 表示将一个内容为空的同名分支推送到远程的分支.(说白 ...

  6. pt-archive提速的实践经验

    最近遇到很多业务需求,需要进行数据导出工作,由于有格式要求,故之前一直使用mysqldump的方法. mysqldump -uuser -ppassword -S mysql.sock -t db t ...

  7. Druid 配置_DruidDataSource参考配置

    以下是一个参考的连接池配置: <bean id="dataSource" class="com.alibaba.druid.pool.DruidDataSource ...

  8. DTCC:MySQl核心代码开发经验揭示

    http://tech.it168.com/a2012/0413/1337/000001337236.shtml

  9. word2010图片仅仅显示边框

    有两个可能的原因: 1.图片所在段落的行间距被设置成固定值了 解决:选择所在段落,右键选择段落--弹出[段落]设置对话框--把固定行距改为"单倍行距"或其它--确定. 2.显示设置 ...

  10. linux 多线程查看工具

    参考: http://www.oschina.net/translate/command-line-tools-to-monitor-linux-performance?cmp&p=1 htt ...