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  • pat1028. List Sorting (25)

    1028. List Sorting (25)

    时间限制
    200 ms
    内存限制
    65536 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CHEN, Yue

    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
    

    提交代码

    scanf printf的速度比cin cout快很多

     1 #include<cstdio>
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<queue>
     6 #include<vector>
     7 #include<cmath>
     8 using namespace std;
     9 struct record{
    10     char name[10];
    11     int id,grade;
    12 };
    13 bool cmp1(record a,record b){
    14     return a.id<b.id;
    15 }
    16 bool cmp2(record a,record b){
    17     if(strcmp(a.name,b.name)==0){
    18         return a.id<b.id;
    19     }
    20     return strcmp(a.name,b.name)<0;
    21 }
    22 bool cmp3(record a,record b){
    23     if(a.grade==b.grade){
    24         return a.id<b.id;
    25     }
    26     return a.grade<b.grade;
    27 }
    28 int main(){
    29     //freopen("D:\input.txt","r",stdin);
    30     int n,c;
    31     scanf("%d %d",&n,&c);
    32     record *r=new record[n+5];
    33     int i;
    34     for(i=0;i<n;i++){
    35         scanf("%d %s %d",&r[i].id,r[i].name,&r[i].grade);
    36         //cin>>r[i].id>>r[i].name>>r[i].grade;
    37     }    
    38     if(c==1){
    39         sort(r,r+n,cmp1);
    40     }
    41     else  if(c==2){
    42         sort(r,r+n,cmp2);
    43     }
    44     else{
    45         sort(r,r+n,cmp3);
    46     }
    47     for(i=0;i<n;i++){
    48         printf("%06d %s %d
    ",r[i].id,r[i].name,r[i].grade);
    49     }
    50 delete []
    r;
    51   return 0; 
    51 }
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  • 原文地址:https://www.cnblogs.com/Deribs4/p/4767708.html
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