PAT甲级1028. List Sorting (25)
来源:互联网 发布:mac安装win10不保留ox 编辑:程序博客网 时间:2024/06/08 06:16
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
#include <cstdio>#include <vector>#include <algorithm>#include <cstring>using namespace std;struct Data{ int id; char name[9]; //需要大于8 int score;};bool cmp1(Data x,Data y){ return x.id<y.id;}bool cmp2(Data x,Data y){ if(strcmp(x.name,y.name)!=0) return strcmp(x.name,y.name)<0; else return x.id<y.id;}bool cmp3(Data x,Data y){ if(x.score!=y.score) return x.score<y.score; else return x.id<y.id;}int main(){ int N,C; scanf("%d %d",&N,&C); vector<Data> Record(N); for(int i=0;i<N;i++){ scanf("%d %s %d",&Record[i].id,&Record[i].name,&Record[i].score); } if(C==1){ sort(Record.begin(),Record.end(),cmp1); for(int i=0;i<N;i++) printf("%06d %s %d\n",Record[i].id,Record[i].name,Record[i].score); } else if(C==2){ sort(Record.begin(),Record.end(),cmp2); for(int i=0;i<N;i++) printf("%06d %s %d\n",Record[i].id,Record[i].name,Record[i].score); } else{ sort(Record.begin(),Record.end(),cmp3); for(int i=0;i<N;i++) printf("%06d %s %d\n",Record[i].id,Record[i].name,Record[i].score); } return 0;}
- 【PAT甲级】1028. List Sorting (25)
- PAT(甲级)1028. List Sorting (25)
- 1028. List Sorting (25) PAT甲级
- PAT甲级练习1028. List Sorting (25)
- PAT甲级1028. List Sorting (25)
- 1028. List Sorting (25)-PAT甲级
- PAT甲级 1028. List Sorting (25)
- PAT甲级 1028. List Sorting (25)
- PAT 甲级 1028. List Sorting (25)
- PAT甲级 1028. List Sorting (25)
- 1028. List Sorting (25)-PAT甲级真题
- 【PAT甲级】1052. Linked List Sorting (25)
- 1052. Linked List Sorting (25) PAT甲级
- PAT甲级1052. Linked List Sorting (25)
- PAT 甲级 1052. Linked List Sorting (25)
- PAT甲级 1052. Linked List Sorting (25)
- PAT 甲级 1041 Linked List Sorting (25)
- 1052. Linked List Sorting (25)-PAT甲级真题
- 渠道对账及差错处理系统设计
- C# ArrayList 的Sort函数使用
- 【转】Ubuntu互换ctrl和caps lock按键位置
- 详解第三方支付之关于会计对账
- Linux下安装配置vsftp
- PAT甲级1028. List Sorting (25)
- java之String, inputStream与Reader转换
- Unity3D 初涉门径之物体移动(W,A,S,D)实现
- map
- Spark性能优化:数据倾斜调优
- c++中的内存对齐
- hello测试博文标题
- 329. Longest Increasing Path in a Matrix
- POJ 3624 (01背包一维求解)