Excel can sort records according to any column. Now you are supposed to imitate this function.
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 Nlines 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).
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.
题目大意：根据c的值是1还是2还是3，对相应的列排序。第一列升序，第二列不降序，第三列不降序。分析：注意，按照名称的不降序排序，因为strcmp比较的是ACSII码，所以A < Z。写cmp函数的时候return strcmp(a.name, b.name) <= 0; return语句返回的是true或者false的值，所以要写 <= 0 这样的形式。比较ACSII码的大小，strcmp(‘a’, ‘z’)返回负值，因为a<z a – z < 0 按照分数的不降序排序，a.score <= b.score~~~
#include <iostream> #include <algorithm> #include <string.h> using namespace std; const int maxn = 100001; struct NODE { int no, score; char name[10]; }node[maxn]; int c; int cmp1(NODE a, NODE b) { if(c == 1) { return a.no < b.no; } else if(c == 2) { if( strcmp(a.name, b.name) == 0) return a.no < b.no; return strcmp(a.name, b.name) <= 0; } else { if(a.score == b.score) return a.no < b.no; return a.score <= b.score; } } int main() { int n; scanf("%d%d", &n, &c); for(int i = 0; i < n; i++) scanf("%d %s %d", &node[i].no, node[i].name, &node[i].score); sort(node, node + n, cmp1); for(int i = 0; i < n; i++) printf("d %s %d\n", node[i].no, node[i].name, node[i].score); return 0; }
